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XI – CLASS WORK PROBLEMS

Q-1) (((( ))))8

79 – + 4 – 3

2

∫∫∫∫

xx dx

Ans. Let I = ( )8

79 – + 4 – 32

xx dx

∫

=( )

××××

9 –4 – 3 12

+ +1 8 4

9 –2

x

xc

9

8

=( )

4 – 32– 9 – + +9 2 32

xxc

89

Q-2)2sin 3∫∫∫∫ x dx

Ans. I =2sin 3x dx∫

=1 – cos6

2

xdx∫

=1

– cos62

dx x dx ∫ ∫

=1 sin6

– +2 6

xx c

=sin6– +

2 12

x xc

Q-4) sin

1+ sin∫∫∫∫x

dxx

Ans. Let I =sin

1+sin

xdx

x∫=

( )

( ) ( )

sin 1 – sin

1+sin 1 – sin

x xdx

x x∫

=( )

2

sin 1 – sin

1 – sin

x xdx

x∫=

( )2

sin 1 – sin

cos

x xdx

x∫= ( )2tan .sec – tanx x x dx∫= ( )2tan .sec – sec –1x x x dx

∫= ( )2tan .sec – sec +1x x x dx∫= secx – tanx + x + c.

Q-3) (((( ))))2

tan3 cot3++++∫∫∫∫ x x dx

Ans. I = ( )2

tan3 + cot3x x dx∫= ( )2 2tan 3 +2tan3 cot3 + cot 3x x x x dx∫= ( )2 2sec 3 –1+2+ cos 3 –1x ec x dx∫= ( )2 2sec 3 + cos 3x ec x dx∫=

2 2sec 3 + cosec 3x x dx∫ ∫Indefinite Integration

= tan3 cot 3

– +3 3

x xc

Q-5) 2 2

1

sin .cos∫∫∫∫ dxx x

Ans. Let I = 2 2

1

sin .cosdx

x x∫=

2 2

2 2

sin + cos

sin .cos

x xdx

x x∫

=

2 2

2 2 2 2

sin cos+

sin .cos sin .cos

x xdx

x x x x

∫

= 2 2

1 1+

cos sindx

x x

∫

= ( )2 2sec + cosx ec x dx∫= tanx – cotx + c

Indefinite Integration

2 Mahesh Tutorials Science

Q-8) If f ′′′′(x) = 4x3 – 2x + 1 and f(2) = 17, find f(x)

Ans. = ( )'f x dx∫= ( )34 – 2 +1x x dx∫=4 2

– + +4 2

x xx c

4 2

Q-9)

26 +17 +10

2 +3∫∫∫∫x x

dxx

Ans. Let I =

26 +17 +10

2 + 3

x xdx

x∫

2

2 ±±±±

±±±±

3 + 4

2 +3 6 +17 +10

–6 9

8 10

–8 12

–2

x

x x x

x x

x

x

+

∴∴∴∴ I = ( )( )

( )

–23 + 4 +

2 + 3x dx

x

∫

=

2

××××3 1

+ 4 – 2log 2 +3 +2 2

xx x c

=

23+ 4 – log 2 + 3 +

2

xx x c

= x2 + 3x +7 log |7x – 1|+ c

Q-10)6 – 8

5 –1∫∫∫∫x

dxx

Ans. I = 6 – 8

5 –1

xdx

x∫Let 6x – 8 = a (5x – 1) + b

6x – 8 = 5ax – (a – b)

Comparing co-efficients

5a = 6 a – b = 8

a = 6

5

6

5– b = 8 ∴∴∴∴ b =

–34

5

I = ( )

( )

6 345 –1 –

5 5

5 –1

x

dxx∫

= ( )

6 34–

5 5 5 –1dx

x

∫

=6 34 1

–5 5 5 –1

dx dxx∫ ∫

=××××

6 34– log 5 –1 +

5 5 5x x c

=6 34– log 5 –1 +

5 25

xx c

Q-6) cos – 4 tan

4 cos∫∫∫∫x x

x

e x xdx

x

Ans. Let I =cos – 4 tan

4 cos

e x xdx

x∫x x

x

=cos 4 tan

–4 cos 4 cos

e x xdx

x x

∫

x x

x x

= – tan .sec4

ex x dx

∫

x

=4

– sec +

log4

e

x ce

x

=( )

– sec +4 log – log 4

ex c

e

x

x

=( )

– sec +4 1 – log 4

ex c

x

x

Q-7)cos2 – cos2

cos – cos

αααα

αααα∫∫∫∫x

dxx

Ans. =( ) ( )2 2 αααα

αααα

2cos –1 – 2cos –1

cos – cos

xdx

x∫=

2 2αααα

αααα

2cos –1 – 2cos +1

cos – cos

xdx

x∫

=( )2 2 αααα

αααα

2 cos – cos

cos – cos

xdx

x∫=

( ) ( )α αα αα αα α

αααα

cos – cos cos + cos2

cos – cos

x xdx

x∫= ( )αααα2 cos + cosx dx∫= αααα2 cos + cos 1x dx dx∫ ∫= 2.sin x + 2 cos αααα × × × × x + c

= 2 sin x + 2x cos α α α α + c

∴∴∴∴ f(x) = x4 – x2 + x + c

∴∴∴∴ f(x) = (2)4 – (2)2 + 2 + c

= 16 – 4 + 2 + c

∴∴∴∴ f(x) = x4 – x2 + x + 3

Mahesh Tutorials Science 3


XI – HOME WORK PROBLEMS

Q-1)

3 2+ 3 + +2

∫∫∫∫x x x

dxx

Ans. Let I =

3 2+3 + +2x x xdx

x∫=

2 2+3 +1+x x dx

x

∫

=3+ + +2log +

3 2

x xx x c

3 2

Q-2)

3 +2 + 5

∫∫∫∫x x

dxx

Ans. Let I =

3 +2 +5x xdx

x∫

=

2 2 5+ +

x xdx

x x x

∫

=

3 1 –1

2 2 2+ 2 +5x x x dx ∫

=

5 3 1

2 2 22 5+ + +

5 3 1

2 2 2

x x xc

=

5 3 1

2 2 22 4

+ +10 +5 3x x x c

Q-3)1

3 – 4∫∫∫∫ dxx

Ans. Let I =1

3 – 4dx

x∫

= ( )

–1

2

3 – 4x dx∫

=

( )

1

2

××××3 – 4 1

+1 –4

2

xc

= ( )

1

21– 3 – 4 +2

x c

Q-4)3cos∫∫∫∫ x dx

Ans. Let I =3cos x dx∫

=cos3 +3cos

4

x xdx∫

=1

cos3 + 3cos4

x dx x dx ∫ ∫

= ( )1 sin3

+ 3 sin +4 3

xx c

∴∴∴∴ I =1 sin3

+3sin +4 3

xx c

Q-5) (((( ))))

3

2

6 +5∫∫∫∫ x dx

Ans. Let I = ( )

3

2

6 +5x dx∫

=

( )

5

2

××××6 +5 1

+5 6

2

xc

= ( )

5

216 +5 +

15x c∫

Q-6) (((( )))) (((( ))))4 7

3 +5 + 5 – 3 ∫∫∫∫ x x dx

Ans. Let I = ( ) ( )4 7

3 +5 + 5 – 3x x dx ∫

=( ) ( )

5 8

× ×× ×× ×× ×3 +5 5 – 31 1

+ +5 3 8 –3

x xc

= ( ) ( )5 81 1

3 +5 – 5 – 3 +15 24

x x c

Q-7)+ 4 –∫∫∫∫dx

x x

Ans. Let I =+ 4 –

dx

x x∫

= ( )( )( )

××××+ 4 +

+ 4 – + 4 +

x xdx

x x x x∫

= ( )

+ 4 –

+ 4 –

x xdx

x x∫



Q-12)1

1 – cosdx

x∫∫∫∫Ans. Let I =

1

1 – cosdx

x∫=

( )

( ) ( )

1+ cos

1 – cos 1+ cos

xdx

x x∫

=( )

2

1+ cos

1 – cos

xdx

x∫= 2

1+ cos

sin

xdx

x∫= ( )2cosec + cot .cosecx x x dx∫= –cot x – cosec x + c

Q-13)cos2

1+ cos2∫∫∫∫x

dxx

Ans. Let I =cos2

1+ cos2

xdx

x∫=

2

2

2cos –1

2cos

xdx

x∫= 2

11 –2cos

dxx

∫

=21

1 – sec2

dx x dx∫ ∫= x –

1

2 tan x + c

Q-9)2cos∫∫∫∫ x xdx

Ans. Let I =2cos x x dx∫

=1+ cos2

2

xdx∫

=1

1 + cos22

dx x dx ∫ ∫

= 1 sin2

+ +4 2

xx c

=1 sin2

+ +2 2

xx c

Q-10) sin5 .cos3∫∫∫∫ x xdx

Ans. Let I = sin5 .cos3x x dx∫=

12sin5 .cos3

2x x dx∫

= ( ) ( )1

sin 5 + 3 +sin 5 – 32

x x x x dx ∫= ( )1

sin8 +sin24

x x dx∫=1 – cos8 cos2

– +2 8 2

x xc

Q-11) 2

4 – 7cos3

sin 3

xdx

x∫∫∫∫Ans. Let I = 2

4 – 7cos3

sin 3

xdx

x∫= 2 2

4 7cos3–

sin 3 sin 3

xdx

x x

∫

=24 cosec 3 – 7x dx∫

cot3 .cosec3x x dx∫=

( )– cos 34cot3– – 7 +

3 3

ec xxc

=4 7– cot 3 + cosec3 +3 3

x x c

= ( )1

+4 +4

x x dx∫

=( )

33

22+ 41

+ +3 34

2 2

x xc

= ( )

33

22

1+ 4 + +

6x x c

Q-8)2cot∫∫∫∫ xdx

Ans. Let I =2cot x dx∫

= ( )2cosec –1x dx∫= – cot x – x + c



Q-14) –1 cos2

tan1+ sin

xdx

x

∫∫∫∫

Ans. Let I =–1 cos2

tan1+sin

xdx

x

∫

=

2 2

–1

2

cos – sin2 2tan

cos + sin2 2

x x

dxx x

∫

=–1

cos – sin2 2tan

cos + sin2 2

x x

dxx x

∫

=–1

1 – tan2tan

1+ tan2

x

dxx

∫

= ( )–1 –1tan 1 – tan tan2

xdx

∫

=ππππ–4 2

xdx

∫

=2ππππ

– +4 4

xx c

Q-15)If f′′′′(x) = 4x3 – 3x2 + 2x + k and f (0) = 1,

f (1) = 4, find f(x).

Ans. By the definition of integral,

f′(x) = ( ) ( )' = 4 – 3 + 2 +f x dx x x x k dx∫ ∫ 3 2

= 4 – 3 + 2 + 1x dx x dx x dx k dx∫ ∫ ∫ ∫3 2

= 4 – 3 + 2 + +4 3 2

x x xkx c

4 3 2

∴∴∴∴ f(x) = x4 – x3 + x2 + kx + c ... (i)

Now f(0) = 1 gives

f(0) = 0 + 0 + 0 + 0 + c = 1

∴∴∴∴ c = 1

∴∴∴∴ from (i), f(x) = x4 – x3 + x2 + kx + 1 ... (ii)

Further f(1) = 4 gives

f(1) = 1 – 1 + 1 + k + 1 = 4

∴∴∴∴ k = 2

∴∴∴∴ from (ii), f(x) = x4 – x3 + x2 + 2x + 1.

Q-16)sin + 5 cot

5 sin∫∫∫∫x x

x

e x xdx

x

Ans. Let I =sin +5 cot

5 sin

e x xdx

x∫x x

x

= + cot .cosec5

ex x dx

∫

x

x

= + cot .cosec5

edx x x dx

∫ ∫

x

=

5– cosec +

log5

e

x ce

∫

x

= ( )– cosec +

5 log – log5

ex c

e∫x

x

= ( )– cosec +

5 1 – log5

ex c∫

x

x

Q-17)

4 2– 3 + 4

∫∫∫∫x x x

dxx

Ans. Let I =

4 2– 3 + 4x x xdx

x∫

=

4 23 4– +

x x xdx

x x x

∫

= ( )3 – 3 + 4x x dx∫=

4 23– + 4 +

4 2

x xx c

Q-18) (((( )))) (((( ))))2 + 4 – 3∫∫∫∫ x x dx

Ans. Let I = ( ) ( )2 + 4 – 3x x dx∫= ( )3 2– 3 + 4 –12x x x dx∫=

4 3 23 4– + –12 +

4 3 2

x x xx c

=4

3 2– + 2 –12 +4

xx x x c



Q-24)2tan∫∫∫∫ x dx

Ans. Let I =2tan x dx∫ = ( )2sec –1x dx∫

=2sec –dx dx∫ ∫ = tan x – x + c

Q-25) (((( ))))sin4 – cos6∫∫∫∫ x x dx

Ans. Let I = ( )sin4 – cos6x x dx∫= sin4 – cos6x dx x dx∫ ∫=

cos4 sin6– – +

4 6

x xc

Q-20)

22 +3 – 4

∫∫∫∫x x

dxx

Ans. Let I =

22 +3 – 4x xdx

x∫

∴∴∴∴ I =

22 3 4+ –

x x xdx

x x x

∫

=

3 1 –1

2 2 22 +3 – 4x x x dx ∫

=

5 3 1

2 2 2

2 +3 – 4 +5 3 1

2 2 2

x x xc

=

5 3 1

2 2 24

+2 – 8 +5x x x c

Q-21) (((( ))))5

3 + 7∫∫∫∫ x dx

Ans. Let I = ( )5

3 +7x dx∫=

( )6

××××3 + 7 1

6 3

x

=( )

63 +7

+18

xc∫

Q-22)1

3 – 2∫∫∫∫ dxx

Ans. Let I =1

3 – 2dx

x∫∴∴∴∴ I = ( )

–1

23 – 2x dx∫

Q-23)1

2 +5 – 2 – 3∫∫∫∫ dxx x

Ans. Let I =1

2 +5 – 2 – 3dx

x x∫

= ( ) ( )2 +5 + 2 – 3

2 +5 – 2 – 3 2 +5 + 2 – 3

x xdx

x x x x∫

=( )( ) ( )

2 +5 – 2 – 3

2 +5 – 2 – 3

x xdx

x x∫

=1

2 +5 + 2 – 38

x x dx ∫

= ( ) ( )

3 3

2 2

× ×× ×× ×× ×

2 +5 2 – 31+ +

3 382 2

2 2

x xc

= ( ) ( )

3 3

2 2

2 +5 2 – 31+ +

8 3 3

x xc

= ( ) ( )

3 3

2 212 +5 + 2 – 3 +

24x x c

Q-19) (((( ))))

6 414 – 5 + 7

– 2cot .cos + 5

∫∫∫∫

x x xdx

x e x

Ans. Let I =( )

6 4– +

+

14 5 7

– 2cot .cosec 5

x x xdx

x x x

∫

=

3

7 5 2

14 – 5 + 737 5

2

x x x

( )–2 –cosec +5 +x x c

= 2x7 – x5 +

3

214

3x + 2 cosec x + 5x + c

= ( )1

2

××××3 – 2 1

+1 –2

2

xc

= ( )1

23 – 2 +x c



Q-26)3sin∫∫∫∫ xdx

Ans. Let I =3sin x dx∫

∵∵∵∵ sin 3x = 3sinx – 4sin3x

⇒⇒⇒⇒ sin3x = 3sin x –– sin3

4

x

=1

4[3sinx – sin 3x]

∴∴∴∴ I =1

4[3sinx – sin 3x] dx

=

1 cos3–3cos + +

4 3

xx c

∴∴∴∴ I =3 cos3– cos + +4 12

xx c

Q-27) 2

3+ 4 sin5

cos 5∫∫∫∫xdx

x

Ans. Let I = 2

3+ 4sin5

cos 5

xdx

x∫= 2 2

3 4sin5+

cos 5 cos 5

xdx dx

x x∫ ∫=

23 sec 5 +4 tan5 .sec5x dx x x dx∫ ∫=

tan5 4sec53 + +

5 5

x xc

=1

5(3 tan 5x + 4sec 5x) + c

Q-28) 1+ sin∫∫∫∫ xdx

Ans. Let I = 1+sinxdx∫

=

2

cos +sin2 2

x xdx

∫

=

sin – cos2 2+ +

1 1

2 2

x x

c

= 2sin – 2cos +2 2

x xc

= 2 sin – cos +2 2

x xc

Q-29)

3 23 + 7 + 7 +1

3 +4∫∫∫∫x x x

dxx

Ans. Let I =

3 23 + 7 +7 +1

3 + 4

x x xdx

x∫

=

2

3 2

3 2

2

2

±±±±

±±±±

±±±±

+ +1

3 + 4 3 +7 +7 +1

–3 4

3 +7 +1

– 3 4

3 +1

– 3 4

– 3

x x

x x x x

x x

x x

x x

x

x

∴∴∴∴ I = ( )2 –3+ +1 +

3 + 4x x dx

x

∫

= ( )2 1+ +1 – 3

3 +4x x dx dx

x∫ ∫=

3 2 3log 3 4+ + – +

3 2 3

xx xx c

+

=

3 2

+ + – 3log 3 + 4 +3 2

x xx x c

Q-30)

28 +10 – 2

2 +1∫∫∫∫x x

dxx

Ans. I =

28 +10 – 2

2 +1

x xdx

x∫

2

2 2

±±±±

4 +3

2 +1 8 +10 – 2

8 + 4

6 – 2

6 3

– 5

x

x x x

x x

x

x

I = ( ) ( )

( )

2 +1 4 +3 – 5

2 +1

x xdx

x∫

= ( )

54 +3 –

2 +1x dx

x

∫

=

24 5+3 – log 2 +1 +

2 2

xx x c

I =2 5

2 +3 – log 2 +1 +2

x x x c



GROUP (A) – CLASS WORK PROBLEMS

Q-1)(((( ))))sin log x

dxx∫∫∫∫

Ans. Let I =( )sin log x

dxx∫

Put log x = t

∴∴∴∴1dx

x = dt

∴∴∴∴ I = sin t dt∫= – cos t + c

= – cos (log x) + c

Q-2)sin cosxe x dx∫∫∫∫

Ans. Let I = cose x dx∫ sin x

Put sin x = t

∴∴∴∴ cos x dx = dt

∴∴∴∴ I = e dt∫ t

= et + c

∴∴∴∴ I = esin x + c

Q-3) 2

sin

1+ cos

xdx

x∫∫∫∫

Ans. Let I =sin

1+ cos

xdx

x∫ 2

Put, cos x = t – sinx dx = dt

∴∴∴∴ I =–

1+

dt

t∫ 2

= – tan–1 (t) + c

= – tan–1 (cos x) + c

Q-4)3sin .cosx x dx∫∫∫∫

Ans. Let I = sin .cosx x dx∫ 3

Put sin x = t

∴∴∴∴ cos dx = dt

∴∴∴∴ I = t dt∫ 3

= +4

tc

4

=( )sin

+4

xc

4

=1sin +

4x c4

Q-5)4sec .tanx x dx∫∫∫∫

Ans. Let I = sec .tanx x dx∫ 4

Put tan x = t

∴∴∴∴ sec2x dx = dt

∴∴∴∴ I = ( )sec .tan secx x x dx∫ 2 2

= ( ) ( )1+ tan .tan . secx x x dx∫ 2 2

= ( )1+ .t t dt∫ 2

= ( )+t t dt∫ 3

= + +2 4

t tc

2 4

=tan tan

+ +2 4

x xc

2 4

or

I = ( )sec . sec .tanx x x dx∫ 3

Put sec x = t

∴∴∴∴ sec x. tan x dx = dt

∴∴∴∴ I = t dt∫ 3

∴∴∴∴ I = +4

tc

4

∴∴∴∴ I =1

4sec4 x + c

Q-6)(((( ))))–1 2

4

sin

1+

x x

dx

x

∫∫∫∫

Ans. Let I =( )sin

1+

x xdx

x∫–1 2

4

Put sin–1(x2) = t

∴∴∴∴

( )( )

12

1 –

x dx

x2

2

= dt

∴∴∴∴1 –

x dx

x4 =2

dt



∴∴∴∴ I =2

dtt ∫

=1

+2 2

tc

2

∴∴∴∴ I = +4

tc

2

∴∴∴∴ I =( )( )sin

+4

xc

–1 2

Q-7)(((( ))))

(((( ))))

+

+

log 2 – log

2

x xdx

x x∫∫∫∫

Ans. LetI =( )

( )

log + 2 – log

+ 2

x xdx

x x∫Put log(x + 2) – log x = t

∴∴∴∴1 1

–+2

dxx x

= dt

∴∴∴∴ ( )

– – 2

+ 2

x xdx

x x = dt

∴∴∴∴ ( )+ 2

dx

x x =–2

dt

∴∴∴∴ I = –2

dtt ∫ =

1–2

t dt∫

=1

– +2 2

tc

2

=1

–4[log (x + 2) – log x]2 + c

=1

–4[log x – log (x + 2)]2 + c

=1

– log +4 + 2

xc

x

2

Q-8)(((( ))))

(((( ))))2

1

sin∫∫∫∫x

x

e xdx

xe

+

Ans. Let I = ( )

( )2

x

x

1+

sin

e xdx

xe∫Put xex = t

∴ xex + ex = dt

dx

ex (x + 1)dx = dt

Q-9) (((( ))))

1

log log logdx

x x x∫∫∫∫

Ans. Let I = ( )

1

log log logdx

x x x∫

I =

( )

( )

1

log

log log

dxx x

x∫

∴ I =( )( )

( )

log log

log log

dx

dx dxx∫

= log |log(log x| + c

Q-10)

–1 –1+

+

e x

e x

x edx

x e∫∫∫∫

Ans. Let I =

–1 –1e e

e x

+

+

x edx

x e∫Put xe + ex = t

∴ (exe – 1 + ex ) dx = dt

∴ e (xe – 1 + ex – 1 ) dx = dt

e

∴ I =

dt

e

t

=1

elog |t| + c

=1

elog |xe + ex| + c

∴ I = 2

1

sindtt∫

=2cosec t dt∫

= –cot t + c

I = –cot (xex) + c



Q-12) (((( ))))

2 2

sin2 1

sin cos∫∫∫∫x x

dxa x b x

+

+

Ans. Let I =( )

2 2

sin2 1+

sin + cos

x xdx

a x b x∫Put asin2x + bcos2x = t

differentiate w.r.t.x

2a sinx cosx – 2b cosx sinx = dt

dx

a sin2x – b sin2x = dt

dx

(a – b) sin2x = dt

dx

sin2x dx = ( )–

dt

a b

I = ( )

1

–

dt

t a b∫

= ( )

1

–a blog +t c

I = 1

–a b2 2log sin + cos +a x b x c

Q-13)

2

2

–1

1∫∫∫∫x

x

edx

e +

Ans. Let I =

2

2

x

x

–1

+1

edx

e∫

I =

2

2

x

x

x

x

–1

+1

e

e dxe

e

∫

=

–

–

x x

x x

–

+

e edx

e e∫

I = ( )–

–

x x

x x

+

+

de e

dx dxe e∫

∴∴∴∴ I = –x xlog + +e e c

Q-11) –+∫∫∫∫ n

dx

x x

Ans. Let I = –n

n

1=

1++

dxdx

x xx

x

∫ ∫

= +1

n

n +1

x dx

x∫Put xn+1 + 1 = t

∴∴∴∴ (n + 1)xn dx = dt

∴∴∴∴ xn dx = +1

dt

n

∴∴∴∴ I = ( )

1=

+1 +1

dt dt

n t n t∫ ∫

=1

log ++1

t cn

=+1n1

log +1 + .+1

x cn

GROUP (A) – HOME WORK PROBLEMS

Q-1) 2 1

2 . log 2

2

x

x

xdx

x++++

++++

++++∫∫∫∫Ans. Let I = ∫

+2 .log2

+2

xdx

x2 1

x

x ++++

Put x2 + 2x + 1 = t

(2x + 2x + 1 log 2) dx = dt

(x + 2x log 2) dx = 2

dt

∴∴∴∴ I =1

2∫ dtt

∴∴∴∴ I =1

2 ∫ t dt

=1

2 log |t| + c

∴∴∴∴ I = + +1log 2

2x c2 1x ++++

Q-2) cos x

dxx∫∫∫∫

Ans. Let I = cos x

dxx∫

Put x = t

∴∴∴∴ 1dx

x = 2dt



Q-6)(((( ))))

(((( ))))2

+1

cos

x

x

x edx

xe∫∫∫∫

Ans. Let I =( )

( )+1

cos

x edx

xe∫ 2

x

x

Put xex = t

∴∴∴∴ (xex + ex)dx = dt

i.e, (x + 1)ex dx = dt

∴∴∴∴ I =cos

dt

t∫ 2

= sec t dt∫ 2

= tan t + c

= tan(xex) + c

Q-7)

2

61+

xdx

x∫∫∫∫

Ans. Let I =1+

xdx

x∫2

6

=( )1+

xdx

x∫

2

23

Put x3 = t

∴∴∴∴ x2 dx = 3

dt

∴∴∴∴ I =1

31+

dt

t

∫ 2

=1

3 1+

dt

t∫ 2

=1

3tan–1 (t) + c

∴∴∴∴ I =1

3tan–1 (x3) + c

Q-3) (((( ))))4 5cos 2 +x x dx∫∫∫∫Ans. Let I = ( )∫ cos 2+x x dx4 5

Put 2 + x5 = t

∴∴∴∴ 5x4 dx = dt

∴∴∴∴ x4 dx = 5

dt

∴∴∴∴ I = cos5

dtt ∫

=1sin +

5t c

= ( )1sin 2+ +

5x c5

Q-4)1

–

2

11+

xxe dx

x

∫∫∫∫

Ans. Let I =1

1+e dxx

∫

1–

2

xx

Put x – 1

x = t

∴∴∴∴ 1 – 1

–x

2=

dt

dx

∴∴∴∴1

1+ dxx

2= dt

∴∴∴∴ I = e dt∫ t

= et + c

= +e c

1–xx

Q-5)(((( ))))2 –1 3

6

tan

1+

x xdx

x∫∫∫∫

Ans. Let I =( )tan

1+

x xdx

x∫2 –1 3

6

Put tan–1(x3) = t

∴ I = ( )cos 2t dt∫= 2 cos t dt∫= 2 sin t + c

= 2 sin x + c

∴∴∴∴( )

( )1

31+

x dxx

2

23

= dt

∴∴∴∴1+

xdx

x

2

6 =3

dt

∴∴∴∴ I =3

dtt ∫

=1

+3 2

tc

2

∴∴∴∴ I =1

6 [tan–1 (x3)]2 + c



Q-8)

2–1

2

2tan

1

++++++++

++++ ∫∫∫∫ xxax dx

x

Ans. I =

2–1

2

2tan

1

++

+ ∫x

ax dxx

x

Put x + tan–1x = t

11+

+1dx

x2 = dt

+ 2

+1

xdx

x

2

2 = dt

∴∴∴∴ I = ∫a dtt

=log

a

a

t

+ c

∴∴∴∴ I = +log

ac

a

–1tanx x++++

Q-9) 2

cosec .cot

1+ cosec

x xdx

x∫∫∫∫

Ans. Let I =cosec .cot

1+ cosec

x xdx

x∫ 2

Put cosec x = t

∴∴∴∴ – cosec x. cot x dx = dt

∴∴∴∴ cosec x. cot x dx = –dt

∴∴∴∴ I =–

1+

dt

t∫ 2

= –tan–1 (t) + c

= –tan–1 (cosec x) + c

Q-10) 2

+1

2 + 2 +

xdx

x x∫∫∫∫

Ans. Let I = +1

2+2 +

xdx

x x∫ 2

Put 2 + 2x + x2 = t

∴∴∴∴ (2 + 2x) dx = dt

∴∴∴∴ (x + 1) dx = 2

dt

∴∴∴∴ I =1

2

dt

t

∫

=1

2t dt∫–1

2

=1

+12

2

tc

–1

2

Q-11)(((( ))))

72

1+ sin2

+ sin

xdx

x x∫∫∫∫

Ans. Let I = ( )

7

1+sin2

sin+∫ 2

xdx

x x

Put x + sin2x = t

(1 + 2sinx cos x) dx = dt

(1 + sin 2x) dx = dt

I = ∫1dt

t7

= ∫ t dt– 7

= ∫ +–7 +1

tc

– 7+1

= ∫ +–6

tc

– 6

I =( )+sin

+–6

x xc

–62

= +t c1

2

= 2+2 + +x x c2

Q-12)1

++++∫∫∫∫ dxx x

Ans. Let I =1

+dx

x x∫=

( )∫1

+1dx

x x

I =

( )∫2

2 +1dx

x x

f (x) = ( )+1x

f ′′′′(x) =1

2 x

∴∴∴∴ I = 2log +1 +x c



Q-19)(((( ))))–1

2

tan

1+

xdx

x∫∫∫∫

Ans. Let I =( )tan

1+

xdx

x∫–1

2

Put tan–1 x = t

∴∴∴∴1

1+ x2 dx = dt

∴∴∴∴ I = t dt∫ 2

= +3

tc

3

=( )tan

+3

xc

3–1

Q-20)–1 2

1

sin . 1 –dx

x x∫∫∫∫

Ans. Let I =1

sin . 1 –dx

x x∫ –1 2

Put sin–1 x = t

∴∴∴∴1

1 – x2 dx = dt

∴∴∴∴ I =dt

t∫= log|t| + c

= log|sin–1x| + c

Q-13)2+5

x

x

e

e∫∫∫∫Ans. Let f (x) = 2 + 5ex

f′ (x) = 5ex

∴∴∴∴ I =1 5

5 2+5

e

e∫x

x dx

I =1

5 log |2 + 5ex| + c

Q-14)cos – sin

cos + sin

x xdx

x x∫∫∫∫

Ans. Let I =cos – sin

cos + sin

x xdx

x x∫Let f ′(x) = cos x + sin x

∴∴∴∴ f ′(x) = –sin x + cos x = cos x – sin x

∴∴∴∴ I = log |cos x + sin x| + c

Q-15) 2 2

sin2

4 sin + cos

xdx

x x∫∫∫∫

Ans. Let I =sin2

4sin + cos

xdx

x x∫ 2 2

Let f ′(x) = 4 sin2 x + cos2 x

∴∴∴∴ f ′(x) = 8 sin x cos x –2 cos x. sin x

= 4 sin 2x – sin x

= 3 sin 2 x

∴∴∴∴ I =1 3sin2

3 4sin + cos

xdx

x x∫ 2 2

=1

3log|4sin2x + cos2x| + c

Q-17)5sin .cosx x dx∫∫∫∫

Ans. Let I = sin .cosx x dx∫ 5

Put sin x = t


Q-16)cos

sin

xdx


Put sin x = t


∴∴∴∴ I =dt

t∫ = 2 +t c

∴∴∴∴ I = 2 sin +x c

Q-18)5sec .tanx x dx∫∫∫∫

Ans. Let I = sec .tanx x dx∫ 5

= ( )sec . sec tanx x x dx∫ 4

Put sec x = t

∴∴∴∴ sec x. tan x dx = dt

∴∴∴∴ I = t dt∫ 4= +

5

tc

5

∴∴∴∴ I =( )sec

+5

xc

5

I =5sec

+5

xc

∴∴∴∴ I = t dt∫ 5= +

6

tc

6

=( )sin

+6

xc

6

∴∴∴∴ I =( )sin

+6

xc

6



Q-21)2 2+

xdx

x a∫∫∫∫

Ans. Let I =+

xdx

x a∫ 2 2

Put x2 + a2 = t

∴∴∴∴ x dx =2

dt

∴∴∴∴ I =1

2

dt

t

∫ = ( )

12 +

2t c

= +t c = + +x a c2 2

Q-22)

–

–

–

++++∫∫∫∫x x

x x

e e

e e

Ans. I =

–

–

–

+∫e e

e e

x x

x x

f(x) = ex + e–x

f ′′′′(x) = ex – e–x

∴∴∴∴ I = log + +e e c–x x

Q-23) 21

x

x++++∫∫∫∫Ans. I = ∫1+

xdx

x2

I =( )∫2

2 1+

xdx

x2

f (x) = (1 + x2)

f ′′′′(x) = 2 x

∴∴∴∴1

2 log |1 + x2| + c

Q-24)4 2tan .sec .x x dx∫∫∫∫

Ans. Put tan x = t

sec2x dx = dt

∴∴∴∴ I = ∫ t dt4

= +5

tc

5

I =tan

+5

xc

5

Q-25)(((( ))))

3–1

2

sin

1 –

xdx

x∫∫∫∫

Ans. I =( )∫

sin

1 –

xdx

x

3–1

2

Put sin–1 x = t

1

1 –dx

x2 = dt

∴∴∴∴ I = ∫ t dt2

= +4

tc

4

I =1

4 (sin–1 x)4 + c

GROUP (B) – CLASS WORK PROBLEMS

Q-1) (((( ))))

sin

sin –

xdx

x a∫∫∫∫

Ans. Let I = ( )

sin

sin –

xdx

x a∫

=( )

( )

sin – +

sin –

x a adx

x a∫

=( )

( )

sin – +

sin –

x a adx

x a

∫

=

( )

( )

( )

sin – .cos

+cos – .sin

sin –

x a a

x a adx

x a

∫

=( )

( )

sin –cos .

sin –

x aa dx

x a∫

( )

( )

cos –+sin .

sin –

x aa dx

x a∫

= ( )cos . + sin cot –a dx a x a dx∫ ∫= x cos a + sin a log |si n (x – a) + c

Q-2) (((( ))))

(((( ))))

cos +

cos –

x adx

x a∫∫∫∫

Ans. Let I =( )

( )

cos +

cos –

x adx

x a∫

=( )

( )

cos – + 2

cos –

x a adx

x a

∫



GROUP (B) – HOME WORK PROBLEMS

Q-1) (((( ))))

cos

sin +

xdx

x a∫∫∫∫

Ans. Let I = ( )cos

sin +

xdx

x a∫

∴∴∴∴ I =( )( )

( )

cos + –

sin +

x a adx

x a∫∴∴∴∴ I =

( ) ( )

( )

cos + cos +sin + sin

sin +

x a a x a adx

x a∫

∴∴∴∴ I = ( )cos cot + + sin 1a x a dx a dx∫ ∫∴∴∴∴ I = cos a log|sin(x + a)|+ sin a. x + c

∴∴∴∴ I = x .sin a + cos a. log|sin(x + a)|+ c

Q-2)(((( ))))

(((( ))))

sin –

sin ++++∫∫∫∫x a

dxx a

Ans. I =( )

( )

sin –

sin +∫x a

dxx a

=( )

( )

sin – 2

sin

+ +∫

x a adx

x a

=( ) ( )

( )

sin cos2 – cos sin2

sin

+ +

+∫x a a x a a

dxx a

Q-3)(((( )))) (((( ))))

1

cos – .sin –dx

x a x b∫∫∫∫

Ans. Let I = ( ) ( )

1

cos – .sin –dx

x a x b∫

=( )

( )

( ) ( )

1.

cos –

cos –

cos – .sin –

b a

b adx

x a x b∫

= ( )

( ) ( )

( ) ( )

sec – .

cos – – –

cos – .sin –

b a

x a x bdx

x a x b

∫

= ( )

( ) ( )

( ) ( )

( ) ( )

sec –

cos – .cos –

+ sin – .sin –

cos – .sin –

b a

x a x b

x a x bdx

x a x b

∫

= ( )

( )

( )

( )

( )

sec –

cos – sin –+

sin – cos –

b a

x b x adx

x b x a

∫

= ( )

( ) ( )

sec –

cot – + tan –

b a

x b dx x a dx ∫ ∫

= sec(b – a)

[log|sin(x – b) – log|cos(x – a)|] + c

= sec(b – a). log ( )

( )

sin –+

cos –

x bc

x a

Q-4)(((( )))) (((( ))))

1

cos – cos –dx

x a x b∫∫∫∫

Ans. Let I = ( ) ( )

1

cos – cos –dx

x a x b∫

=

( )

( )

( )

cos – .cos2

– sin – .sin2

cos –

x a a

x a adx

x a

∫

= ( )cos2 – tan – sin2a dx x a a dx∫ ∫

= ( )cos2 – sin2 tan –a dx a x a dx∫ ∫= x. cos 2a – sin2 a.log|sec (x – a)|+ c

= ( )

( )

( ) ( )

sin –1

sin – cos – cos –

b adx

b a x a x b∫

= ( )

( ) ( )

( ) ( )

sin – – –1

sin – cos – cos –

x a x bdx

b a x a x b

∫

= ( )

1

sin –b a××××

( ) ( ) ( ) ( )

( ) ( )

sin – cos – – cos – sin –

cos – cos –

x a x b x a x bdx

x a x b∫

= ( )cosec ( – ) tan – – tan( – )b a x a dx x b dx ∫ ∫

= ( )( )

( )

log sec –cosec –

– log sec – +

x ab a

x b c

= ( )( )

( )

sec –cosec – log +

sec –

x ab a c

x b



Q-3)(((( )))) (((( ))))

1

sin – cos –dx

x a x b∫∫∫∫

Ans. Let I = ( ) ( )

1

sin – cos –dx

x a x b∫

∴∴∴∴ I = ( )

( )

( ) ( )

cos –1.

cos – sin – .cos –

b adx

b a x a x b∫

∴∴∴∴ I = ( )

( ) ( )

( ) ( )

cos – – –1

cos – sin – .cos –

x a x bdx

b a x a x b

∫

I = sec(b – a)

( ) ( ) ( ) ( )

( ) ( )

cos – cos – +sin – sin –

sin – .cos –

x a x b x a x bdx

x a x b∫

I = ( ) ( ) ( )sec – cot – + tan –b a x a x b dx ∫I = sec(b – a) [log|sin(x – a)|

+ log|sec(x – b)|] + c

= sec(b – a) [log|sin(x – a)|

– log|cos(x – b)] + c

I = ( )( )

( )

sin –sec – log +

cos –

x ab a c

x b

Q-4) (((( )))) (((( ))))

1

sin – sin –dx

x a x b∫∫∫∫

Ans. = ( ) ( )

1

sin – sin –dx

x a x b∫

= ( )

( )

( ) ( )

sin –1

sin – sin – sin –

a b dx

a b x a x b∫

=( )

( ) ( )

( ) ( )

sin – – –1

sin – sin – sin –

x b x adx

a b x a x b

∫

= ( )

1

sin –a b

( ) ( ) ( ) ( )

( ) ( )

sin – cos – – sin – .cos –

sin – sin –

x b x a x a x bdx

x a x b∫

= ( )

1

sin –a b

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

sin – cos – – sin – .cos –

sin – sin – sin – .sin –

x b x a x a x bdx

x a x b x a x b∫

= ( )( ) ( )

1cot – – cot –

sin –x a x b dx

a b ∫

= ( )

( ) ( )

1

sin –

cot – – cot –

a b

x a dx x b dx ∫ ∫

= ( )

( ) ( )

1

sin

log sin – – log sin – + .

a b

x a x b c

−

= ( )

( )

( )

sin –1log +

sin – sin –

x ac

a b x b

Q-5)(((( ))))cos

cos

++++

∫∫∫∫x a

dxx

Ans. I =( )cos

cos

+

∫x a

dxx

=cos .cos – sin sin

cos∫x a x a

dxx

= [ ]cos – tan sin∫ a x a dx

= cos – tan sin∫ ∫adx x adx

= cos – sin tan∫ ∫a dx a x dx

= cos a (x) – sin a log |sec x| + c

I = x cos a – sin a. log |sec x| + c

Q-6) (((( ))))sin +

sin

x adx

x∫∫∫∫

Ans. Let I =( )sin +

sin

x adx

x∫

=sin .cos + cos .sin

sin

x a x adx

x∫

= ( )cos + cot .sina x a dx∫

= cos + sin cota dx a x dx∫ ∫= x.cos a + sin a. log|sin x| + c

= ( )cos2 – cot sin2+ ∫ a x a a dx

= ( )cos2 – cos sin2+∫ ∫adx x a adx

= ( )cos2 – sin2 cot +∫ ∫a dx a x a dx

I = x cos 2a – sin 2 a log |sin (x + a)| + c



GROUP (C) – CLASS WORK PROBLEMS

Q-1) (((( )))) (((( ))))6

3 –1 +1x x dx∫∫∫∫

Ans. Let I = ( ) ( )6

3 – 1 +1x x dx∫Put x + 1 = t

∴∴∴∴ x = t – 1

∴∴∴∴ dx = dt

∴∴∴∴ I = ( ) 63 –1 –1t t dt ∫

= ( ) 63 – 4t t dt∫

=7 63 – 4t dt t dt∫ ∫

=

8 73 4– +

8 7

t tc

= ( ) ( )8 73 4

+1 – +1 +8 7x x c

Q-2) 3

2 +3

3 – 2

xdx

x∫∫∫∫

Ans. Let I = 3

2 + 3

3 – 2

xdx

x∫Put 3x – 2 = t

∴∴∴∴ dx =1

3dt

∴∴∴∴ I =3

+ 22 + 3

3

3

t

dt

t

∫

= 1

3

1 2 + 4 + 9

33

tdt

t

∫

=

2 –1

3 31

2 +139

t t dt ∫

=

5 –1

3 31 2 13+ +

5 29

3 3

t tc

∫

= ( ) ( )

5 2

3 31 6 393 – 2 + 3 – 2 +

9 5 2x x c

Q-3)5 3 3∫∫∫∫ x a x dx+

Ans. Let I =5 3 3+x a x dx∫

=3 3 3 2+x a x x dx⋅∫

Put a3 + x3 = t

x3 = t –a3

3x2dx = dt

∴∴∴∴ x2dx = 3

dt

∴∴∴∴ I = ( )3–3

dtt a t∫

= 1

3( )3 2 3 1 2–t a t dt∫

= 1

3

5 2 3 3 2

– + c5 3

2 2

t a t

= ( ) ( )

53 2

3 3 3 3 322 + 2 +– +

15 9

a x a a xc

Q-4)(((( ))))

7

241+

∫∫∫∫x

dx

x

Ans. Let I =( )

7

241+

xdx

x∫

=( )

4 3

241+

x xdx

x

⋅

∫

Put 1 + x4 = t ∴∴∴∴ x4 = t – 1

4x3 dx = dt

∴∴∴∴ x3dx = 4

dt

∴∴∴∴ I = 2

1 – 1

4

tdt

t∫

= 2

1 1 1–

4dt

t t

∫

=1 1

log + +4

t ct

∫

=1 1

log + +4 4

t ct



Q-2) (((( ))))

5

22 –x a x dx∫∫∫∫

Ans. Let I = ( )∫ –x a x dx

5

22

Put a – x = t

∴ dx = –dt

∴ I = ( ) ( )∫ – –a t t dt

5

22

= ( )∫– – 2 + .a at t t dt

5

22 2

= ∫– 2 – –at t a t dt

7 9 522 2 2

=

2 – – +9 14 7

2 5 2

t t ta a c

9 17 7

2 5 22

=4 5 2

– – +9 14 7

aat t t c

9 14 72

2 5 2

= ( ) ( )

( )

4 2– – –

9 7

2– – +

11

aa a x a x

a x c

9 72

2 2

11

2

Q-3) 2 + 3

4 – 5

xdx

x∫∫∫∫

Ans. Let I =2 + 3

4 – 5

xdx

x∫Put 4x – 5 = t

∴ dx = 4

dt

GROUP (C) – HOME WORK PROBLEMS

Q-1) (((( )))) (((( ))))5

2 –1 + 4x x dx∫∫∫∫

Ans. Let I = ( ) ( )∫ 2 –1 + 4x x dx5

Put x + 4 = t

∴ dx = dt

∴ I = ( ) ∫ 2 – 4 –1t t dt5

= ( )∫ 2 – 8 –1t t dt5

Q-5)

3

3+

xdx

x x∫∫∫∫

Ans. Let I = ∫ +

xdx

x x

3

3

Put x = t6

∴ dx = 6t5 dt

∴ I = ( )∫ 6+

tt dt

t t

25

3 2

∴ I = ∫6 +1

tdt

t

2

= ( )( )

∫

–16 – + + +1 +

+1t t t t dt

t

4 3 2

= ( )( )

∫ ∫

–16 – + + +1 –

+1t t t t dt dt

t

4 3 2

= ∫6 – + – + – log +1 +

5 4 3 2

t t t tt t c

5 4 3 2

∴ I =

1 1 1 1– + –

5 4 3 26 +

+ – log 1+

x x x x

c

x x

5 2 1 1

6 3 2 3

1 1

6 6

= ( )( )

4

4

1 1log 1+ + +

4 4 1+x c

x

= ( )( )

4

4

1 1log 1+ + +

4 1+x c

x

= ( )∫ 2 – 9t t dt5

= ( )∫ 2 – 9t t dt6 5

=

2 – 9 +7 6

t tc

7 6

= ( ) ( )2 3

+ 4 – + 4 +7 2x x c

7 6



Q-4) 4

– 5

2 –1

xdx

x∫∫∫∫

Ans. Let I = 4

– 5

2 –1

xdx

x∫Put 2x – 1 = t

∴∴∴∴ dx = 2

dt

∴∴∴∴ I =4

+1– 5

2

2

t

dt

t

∫

∴∴∴∴ I =

∫

1 – 9

4

tdt

t

1

4

= ∫

1– 9

4t t dt

3 –1

4 4

=

∫

1– 9 +

7 34

4 4

t tc

7 3

4 4

=

1 4–12 +

4 7t t c

7 3

4 4

= ( ) ( )

1 42 –1 –12 2 –1 +

4 7x x c

7 3

4 4

∴∴∴∴ I = ( ) ( )12 –1 – 3 2 –1 +

7x x c

7 3

4 4

Q-5) (((( ))))

4

2 –1

– 5

xdx

x∫∫∫∫

Ans. Let I =( )∫2 1

– 5

xdx

x4

–

Put x – 5 = t

∴∴∴∴ x = t + 5

∴∴∴∴ dx = dt

∴∴∴∴ I =( ) ∫2 +5 –1t

dtt4

= ∫2 +10 –1t

dtt4

= ∫2 +9t

dtt4

= ∫2 9+ dt

t t3 4

= 2 + 9 +–2 –3

t tc

–2 –3

= –(x – 5)–2 – 3(x – 5)–3 + c

∴∴∴∴ x = +5

4

t

∴∴∴∴ I =

+52 + 3

4

4

t

dt

t

∫

=1 +5 + 6

4 2

tdt

t

∫

=

∫

1+11

8t t dt

1 –1

2 2

=

1+ 11 +

3 18

2 2

t tc

3 1

2 2

= ( ) ( )

1 24 – 5 + 22 4 – 5 +

8 3x x c

3 1

2 2

= ( )

1 114 – 5 + 4 – 5 +

12 4x x c

3

2

Q-6) (((( ))))3 – 2 2 –1x x dx∫∫∫∫

Ans. Let I = ( )3 – 2 2 –1x x dx∫Put 2x – 1 = t

∴∴∴∴ dx = 2

dt

∴∴∴∴ I =( )–1

3 – 22 2

t dtt

∫

= ( )1

3 +3 – 44

t t dt∫



GROUP (D) – CLASS WORK PROBLEMS

Q-1)7 sin + 24cos

3cos + 4 sin

x xdx

x x∫∫∫∫Ans. Let 7 sinx + 24 cos x

= A [3 cosx + 4 sinx] + B [– 3 sinx + 4 cos x]

= (4A – 3B) sin x + (3A + 4B) cos x

∴∴∴∴ 4A – 3B = 7 and 3A + 4B = 24

on solving, we get,

A = 4, B = 3

∴∴∴∴ 7 sin x + 24 cos x = 4 (3 cos x + 4 sinx)

+ 3(4cosx – 3sinx)

∴∴∴∴ I =( )

( )

4 3cos + 4sin+

3cos + 4sin

x xdx

x x∫( )

( )

3 4cos 3sin

3cos 4sin

x xdx

x x

−

+∫=

4cos – 3sin4 + 3

3cos + 4sin

x xdx dx

x x∫ ∫Let f (x) = 4 sinx + 3 cosx

f′′′′ (x) = 4 cosx – 3 sinx

∴∴∴∴ I = 4x + 3log |3 cos x + 4 sin x|+ c

Q-8)

2

3 3

xdx

x a−−−−∫∫∫∫

Ans. I = ∫ –

xdx

x a

2

3 3

Put x3 – a3 = t

(3x2)dx = dt

x2dx = 3

dt

∴∴∴∴ I =1

3∫dt

t

= ∫1

3t dt–1 2

=1

+3 1 2

tc

–1 2

=2

+3t c

= ( )2

– +3

x a c3 3

Q-7) 1 1

2 3

1

+

dx

x x

∫∫∫∫

Ans. Let I = ∫1

+

dx

x x

1 1

2 3

Put x = t6

∴∴∴∴ dx = 6t5 dt

∴∴∴∴ I = ∫6

+

t dt

t t

5

3 2

= ∫6 +1

tdt

t

3

=( )

( )∫–1 –1

6+1

tdt

t

3

= ∫

+1 16 –

+1 +1

tdt

t t

3

=( ) ( ) ∫

+1 – +1 16 –

+1 +1

t t tdt

t t

2

= ∫

16 – +1 –

+1t t dt

t

2

=

6 – + – log +1 +3 2

t tt t c

3 2

=

1 16 – + – log 1+ +

3 2x x x x c1 1 1 1

2 3 6 6

= ∫

13 –

4t t dt

3 1

2 2

=

13 – +

5 34

2 2

t tc

5 3

2 2

= ( ) ( )

1 6 22 – 1 – 2 – 1 +

4 5 3x x c

5 3

2 2

= ( ) ( )3 1

2 –1 – 2 –1 +10 6

x x c

5 3

2 2



Q-2)8sin cos

2sin – 3cos

x x

x x

++++

∫∫∫∫Ans. Let 8 sin x + cos x

= A[2 sinx – 3 cosx] + B[2 cosx + 3 sinx]

= (2A + 3B) sin x + (– 3A + 2B) cos x

∴∴∴∴ 2A + 3B = 8

– 3A + 2B = 1

On solving,we get,

A = 1, B = 2

∴∴∴∴ Nr. = (2 sinx – 3 cosx) + 2 (2 cosx + 3 sinx)

∴∴∴∴ I =2cos + 3sin

1 + 22sin – 3cos

x xdx dx

x x∫ ∫= x + 2 log|2 sin x – 3 cos x| + c

Q-3)1

1+ cotdx

x∫∫∫∫Ans. I =

1

1+ cotdx

x∫=

1

cos1+

sin

dxx

x

∫

=sin

sin + cos

xdx

x x∫=

1 2sin

2 sin + cos

xdx

x x∫

=( ) ( )( )

( )

sin + cos + sin – cos1

2 sin + cos

x x x xdx

x x∫

=( )

( )

cos – sin11. –

2 sin + cos

x xdx dx

x x

∫ ∫

=1

– log sin + cos +2x x x c

Q-4)3 – 4

4 +5

x

x

edx

e∫∫∫∫Ans. Let I = ∫

3 – 4

4 + 5

edx

e

x

x

Let 3ex – 4 = A (4ex + 5) + B d

dx(4ex + 5)

= A(4e2x + 5) + B (4ex )

= ex (4A + 4B) + 5A

∴∴∴∴ 4A + 4B = 3 and

– 5A = –4

∴∴∴∴ A =4

–5

GROUP (D) – HOME WORK PROBLEMS

Q-1) 14 sin +8cos

2sin +3cos

x xdx

x x∫∫∫∫Ans. Let 14 sin x + 8 cos x

= A(2 sin x + 3 cos x)

+ Bd

dx (2 sin x + 3 cos x)

= A(2 sin x + 3 cos x) + B(2 cos x – 3 sin x)

= (2A – 3B) sin x + (3A + 2B) cos x

2A – 3B = 14 ...(i)

3A + 2B = 8 ...(ii)

Solving (i) and (ii) we get

A = 4 and B = –2

14 sin x + 8 cos x

= 4(2 sin x + 3 cos x) –2(2 cos x – 3 sin x)

So,

I =( ) ( )4 2sin + 3cos – 2 2cos – 3sin

2sin + 3cos

x x x xdx

x x∫

∴∴∴∴ I =2cos – 3sin

4 – 22sin + 3cos

x xdx

x x

∫

∴∴∴∴ I = 4x – 2 log|2 sin x + 3 cos x| + c

∴∴∴∴ B =3 4 31

+ =4 5 20

∴∴∴∴ I =

( ) ( )

( )∫4 31

– 4 + 5 + 4 + 55 20

4 + 5

de e

dx

e

x x

x

=( )( )

( )

( )∫ ∫4 + 54 + 54 31

– + .5 204 + 5 4 + 5

dee

dxdx dxe e

xx

x x

= ∫4 31

– + log 4 + 5 +5 20

dx e cx

=4 31

– + log 4 + 5 +5 20x e cx



Q-4)20 –12

3 – 4

x

x

edx

e∫∫∫∫

Ans. = ∫10 – 6

23 – 4

edx

e

x

x

=( )∫

6 –10–2

3 – 4

edx

e

x

x

= ∫6 – 8 2

–2 –3 – 4 3 – 4

edx

e e

x

x x

= ∫ ∫

2–2 2 –

3 – 4

edx dx

e

–

–

x

x

Let 3 – 4e–x = f (x)

∴∴∴∴ f′ (x) = 4e–x

= ∫ ∫

1 4–2 2 –

2 3 – 4

edx dx

e

–

–

x

x

=

1–2 2 – log 3 – 4 +

2x e c–x

= log |3 – 4e–x| – 4x + c

= log |3ex – 4| – log ex – 4x + c

= log |3ex – 4| – x log e – 4x + c

= log |3ex – 4| – 5x + c ..... (∵∵∵∵ log e = 1)

Q-3)3 + 4

2 – 8

x

x

edx

e∫∫∫∫

Ans. Let I = ∫3 + 4

2 – 8

edx

e

x

x

Let 3ex + 4 = A(2ex – 8) + Bd

dx(2ex – 8)

∴∴∴∴ 3ex + 4 = A(2ex – 8) + B(2ex)

∴∴∴∴ 3ex + 4 = (2A + 2B) ex – 8A

∴∴∴∴ 2A + 2B = 3 ...(i)

–8A = 4 ...(ii)

Solving (i) and (ii) we get

A = 1

–2

and B = 2

∴∴∴∴ 3ex + 4 = 1

–2

(2x – 8) + 2(2ex)

I =( ) ( )

∫1

– 2 – 8 + 2 22

2 – 8

e e

dxe

x x

x

I =( )

∫

21– + 2

2 2 – 8

edx

e

x

x

I =1

–2x + 2log|2ex – 8| + c

Q-2)1

1 tandx

x∫∫∫∫ ++++

Ans. I =1

1+ tandx

x∫=

1

sin1+

cos

dxx

x

∫

=cos

cos + sin

xdx

x x∫=

1 2cos

2 cos + sin

xdx

x x∫

=( ) ( )( )

( )

cos + sin + cos sin1

2 cos + sin

x x x xdx

x x

−

∫

=( )

( )

cos – sin11. +

2 sin + cos

x xdx dx

x x

∫ ∫

=1

+ log sin + cos +2x x x c



GROUP (E) – CLASS WORK PROBLEMS

Q-1) 2

1

4 + 9dx

x∫∫∫∫Ans. Let I = ∫

1

4 + 9dx

x2

= ( ) ( )∫1

2 + 3dx

x22

=

1 2 1tan +

3 3 2

xc–1 ××××

=

1 2tan +

6 3

xc–1

Q-2) 2

sec .tan

9 –16 sec

x xdx

x∫∫∫∫

Ans. Let I =sec .tan

9 –16sec

x xdx

x∫ 2

Put sin x = t

∴∴∴∴ sec x . tan x dx = dt

∴∴∴∴ I = ∫ 9 –16dt

t2

= ( ) ( )∫ 3 – 4

dt

t2 2

= ××××××××

1 3+ 4 1log +

2 3 3 – 4 4

tc

t

=1 3+4seclog +

24 3 – 4sec

xc

x

Q-5) 2

1

+6 +10dx

x x∫∫∫∫Ans. Let I = ∫

1

+6 +10dx

x x2

= ∫1

+6 +9+1dx

x x2

Q-3) 2

cos

10 – cos∫∫∫∫x

dx

x

Ans. Let I = 2

cos

10 – cos

xdx

x∫

= ( )2

cos

10 – 1 – sin

xdx

x∫

I = 2

cos

9 + sin

xdx

x∫Put sin x = t

∴∴∴∴ cosx dx = dt

∴∴∴∴ I =( )

2 2

1

3 + tdt∫

=( )

22

1

t + 3dt∫

I =2log t + t + 9 + c

I =2log sin + sin + 9 +x x c

=2log sin + 1 – cos + 9 +x x c

I =2log sin + 10 – cos +x x c

Q-4) –

1

4 + 9e∫∫∫∫ x xdx

e

Ans. Let I = –x x

1

4 + 9edx

e∫

= 2

x

x4 +9

edx

e∫Put ex = t ∴∴∴∴ exdx = dt

I = 2

1

4 + 9dt

t∫

=( )

2 2

1

2 +3dt

t∫

= ( )22

1 1

4 + 3 2dt

t∫

=–11 1

tan +34 3 2

2

tc

=–11 2

tan +6 3

tc

I =–1

x1 2tan +

6 3

ec



Q-9)

2

2

sec

5tan –12 tan +14

xdx

x x∫∫∫∫

Ans. Let I = ∫sec

5 tan –12tan +14

xdx

x x

2

2

Put tan x = t

sec2 x dx = dt

∴∴∴∴ I = ∫ 5 –12 +14

dt

t t2

= ∫1

12 145– +5 5

dt

t t2

= ∫1

12 144 14 1445– + + –5 100 5 100

dt

t t2

=

∫1

5 12 2 34– +10 10

dt

t

22

=

12–

1 1 10tan +5 2 342 34

1010

t

c–1

××××

=

1 10 –12tan +

34 2 34

tc

–1

=

1 5 tan – 6tan +

34 34

xc

–1

Q-6) 2

1

15+4 – 4dx

x x∫∫∫∫

Ans. Let I = ∫1

15 + 4 – 4dx

x x2

=( )∫15 – 4 – 4 +1 +1

dtdx

x x2

=( ) ( )∫ 4 – 2 –1

dtdx

x2 2

= ××××××××

1 4+ 2 –1 1log +

2 4 4 – 2 +1 2

xc

x

=1 3+2log +

16 5 – 2

xc

x

Q-7) 2

1

+8 +25dx

x x∫∫∫∫

Ans. Let I = ∫1

+8 + 25dx

x x2

= ∫1

+ 8 +16 + 9dx

x x2

=( ) ( )∫

1

+ 4 + 3dx

x2 2

= ( ) ( ) ( )log + 4 + + 4 + 3 +x x c2 2

= ( )log + 4 + + 8 + 25 +x x x c2

Q-8) 2

1

3 – 4 +2dx

x x∫∫∫∫

Ans. Let I = ∫1

3 – 4 + 2dx

x x2

=1 1

3 4 2– +3 3

dx

x x∫

2

=1 1

3 4 4 2 4– + + –3 9 3 9

dx

x x∫

2

= ( )∫1

+ 3 +1dx

x2 2

= tan–1 (x + 3) + c

=1 1

32 2

– +3 3

dx

x

∫ 22

=

1log

3

2 2 2– + – – +3 3 3

x x c

22

=1 2 4 2log – + – + +

3 33x x x c

x

2



Q-10) 2 +4 +13

x

x x

edx

e e∫∫∫∫

Ans. Let I = ∫ +4 +13

edx

e e2

x

x x

put ex = t

∴∴∴∴ ex dx = dt

∴∴∴∴ I = ∫ + 4 +13

dt

t t2

= ∫ + 4 + 4 + 9

dt

t t2

=( ) ( )∫ +2 2+ 3

dt

t2

= ( ) ( ) ( )log + 2 + + 2 + 3 +t t c2 2

= ( ) ( ) ( )log + 4 + + 2 + 3 +e e c2x t

= ( )log + 2 + + 4 +13 +e e e cx 2x x

Q-11)2sin cos – 2cos 2x x x dx++++∫∫∫∫

Ans. I = ∫sin cos – 2cos +2x x x dx2

Put cos x = t ∴∴∴∴ –sin x dx = dt

I = ∫– – 2 +2t t dt2

= ( )∫– –1 +1t dt2 2

=

( )

( )

–1–1 +1

2– +

1+ log –1+ –1 +122

tt

c

t t

2 2

2

=

cos –1– cos – 2cos2+2

2

xx2

1+ log cos –1+ cos – 2cos2+ 2 +2

x x c2

=1 – cos

cos – 2cos + 22

xx x2

1

– log cos –1+ cos – 2cos + 2 +2

x x x c2

Q-12) I =2/3 2/3

1

– 4dx

x x∫∫∫∫Ans. I = Put x1/3 = t

∴∴∴∴–2/31

3x dx dt=

∴∴∴∴ –2/3 3x dx dt=

I =2 2

13

– 2dt

t∫=

2 23 log + – 2 +t t c

I =1/3 2/33 log + – 4 +x x c

Q-13) I =sin cos

7 – 9sin2

++++

∫∫∫∫x x

dxx

Ans. Let I = –9

sec + cos

7 sin2

x xdx

x∫Put t = sin x – cos x

∴∴∴∴ dt = (cos x + sin x)dx

and t2 = sin2x + cos2x – 2 sin x.cos x

∴∴∴∴ t2 = 1 – sin 2x

∴∴∴∴ sin 2x = 1 – t2

∴∴∴∴ I = ( )27 – 9 1 –

dt

t∫

= 27 – 9+ 9

dt

t∫= 29 – 2

dt

t∫

= ( ) ( )22

3 – 2

dt

t∫

= ××××1 3 – 2 1

log +32 2 3 + 2

tc

t

∴∴∴∴ I =( )

( )

3 sin – cos – 21log +

6 2 3 sin – cos + 2

x xc

x x



GROUP (E) – HOME WORK PROBLEMS

Q-1) 2

1

+6 +25dx

x x∫∫∫∫

Ans. Let I = 2

1

+6 +25dx

x x∫= 2

1

+6 +9+16dx

x x∫

=( ) ( )

2 2

1

+3 + 4dx

x∫

=–11 +3

tan +4 4

xc

Q-2) 2

1

9 +6 +5dx

x x∫∫∫∫

Ans. Let I = 2

1

9 +6 +5dx

x x∫=

2

1 1

2 59– +3 9

dx

x x∫

Q-14) cos + sin

7 – 2cos2

x xdx

x∫∫∫∫Ans. Let I = –2

cos +sin

7 cos2

x xdx

x∫=

cos sin+

7 – 2cos2 7 – 2cos2

x xdx dx

x x∫ ∫

=( )

( )

2

2

cos

7 – 2 1 – 2sin

sin+

7 – 2 2cos –1

xdx

x

xdx

x

∫

∫

= 2 2

cos sin+

5 +4sin 9 – 4cos

x xdx dx

x x∫ ∫Put sin x = t

1, in first integral

∴∴∴∴ cos x dx = dt1 and

cos x dx = t2 in second integral

∴∴∴∴ sin x dx = –dt2

∴∴∴∴ I =21

2 21 2

–+

5 + 4 9 – 4

dtdt

t t∫ ∫

=( ) ( ) ( ) ( )

21

2 22221

–3 – 25 + 2

dtdt

tt∫ ∫

=–1 1

2

2

××××

××××××××

21 1tan

25 5

3+ 21 1– log +2 3 3 – 2 2

t

tc

t

∴∴∴∴ I =–11 2sin

tan2 5 5

1 3 +2cos– log +16 3 – 2cos

x

xc

x

Q-15)cos – sin

8 – sin2∫∫∫∫x x

dxx

Ans. I =( )

cos – sin

9 – 1 sin2

x xdx

x+∫

= ( )2 2

cos – sin

9 – cos + sin + 2sin cos

x xdx

x x x x∫

=( )

22

cos – sin

3 – cos + sin

x xdx

x x∫Put sin x + cos x = t

∴∴∴∴ (cos x – sin x) dx = dt

Q-16) cos cos2∫∫∫∫ x x dx

Ans. I =2cos 1 – 2sinx x dx∫

Put sin x = t cos x dx = dt

I =21– 2t dt∫

=21

4 –2

t dt∫

=

2

214 –

2t dt

∫

= 2 –11 1/ 2

4 – + sin +2 2 2 1/ 2

t tt c

I = ( )–1sin 1cos2 + sin 2sin +

2 2 2

xx x c

∴∴∴∴ I =–1

2 2

1= sin +

33 –

tdt c

t

∫

∴∴∴∴ I =–1 sin + cos

sin +3

x xc



Q-3) 2

1

9+8 +dx

x x∫∫∫∫

Ans. Let I = 2

1

9+8 +dx

x x∫

= ( )2

1

9 – – 8 +16 +16x x∫

= ( ) ( )2 2

1

5 – – 4dx

x∫

=–1 – 4

sin +5

xc

Q-4) 2

1

2 – 4 +7dx

x x∫∫∫∫Ans. Let I =

2

1

2 – 4 + 7dx

x x∫=

1 1

2 7– 2 +

2

dx

x x∫

2

=1 1

2 7– 2 +1+ –1

2

dx

x x∫ 2

=

( )

1 1

25

–1 +2

dx

x

∫ 2

2

= ( ) ( )

2

21 5log –1 + –1 + +

22x x c

= ( ) 21 7log –1 + – 2 + +

22x x x c

Q-5)

(((( ))))2

1

log + 4dx

x x∫∫∫∫

Ans. Let I =

( )2

1

log + 4dx

x x∫

Put log x = t ∴∴∴∴1

xdx = dt

∴∴∴∴ I =( ) ( )

2 2 2=

+ 4 + 2

dt dt

t t∫ ∫

= ( )22log + + 2 +t t c

= ( )2

log log + log + 4 +x x c

Q-6) 2

cos

sin – 2sin +5

xdx

x x∫∫∫∫

Ans. Let I = 2

cos

sin – 2sin +5

xdx

x x∫Put sin x = t


∴∴∴∴ I =2 – 2 +5

dt

t t∫

=2 – 2 +1+ 4

dt

t t∫

=( ) ( )

2 2–1 + 2

dt

t∫

= ( ) ( ) ( )2 2

log –1 + –1 + 2 +t t c

= ( ) 2log –1 + – 2t + 5 +t t c

= ( ) 2log sin –1 + sin – 2sin +5 +x x x c

=2

1 1

2 1 5 19– + + –3 9 9 9

dx

xx∫

= 2 2

1 1

9 1 2+ +3 3

dx

x

∫

= –1

11 1 3tan +2 29

3 3

x

c

+

=–11 3 +1

tan +6 2

xc

Q-7) 21–

x

x

adx

a∫∫∫∫Ans. put ax = t

ax log a dx = dt

ax dx = log

dt

a



Q-10) (((( )))) (((( ))))2

1

1+ cos2 1 – tandx

x x∫∫∫∫

Ans. Let I =( ) ( )2

1

1+ cos2 1 – tandx

x x∫

= ( ) ( )

2

2 2 2

cos

2cos cos – sin

xdx

x x x∫

=1

sec22

x dx∫=

1log sec2 + tan2 +

4x x c

Q-11) sin + cos

3 +4 sin2

x xdx

x∫∫∫∫

Ans. Let I =sin + cos

3+ 4sin2

x xdx

x∫Put sin x – cos x = t

∴∴∴∴ (cos x + sin x)dx = dt

∴∴∴∴ t2 = sin2x + cos2x – 2 sin x.cos x

t2 = 1 – sin 2x

∴∴∴∴ sin 2x = 1 – t2

∴∴∴∴ I = ( )23+ 4 1 –

dtdx

t∫

= 27 +4

dt

t∫=

( ) ( )2 2

7 – 2

dt

t∫

=( )

××××1 7 + 2 1

log +27 – 22 7

tc

t

∴∴∴∴ I =( )

( )

7 +2 sin – cos1log +

4 7 7 – 2 sin – cos

x xc

x x

Q-12) cos + sin

3 – 2cos2

x xdx

x∫∫∫∫

Ans. Let I =cos +sin

3 – 2cos2

x xdx

x∫=

cos sin+

3 – 2cos2 3 – 2cos2

x xdx dx

x x∫ ∫

Q-8)

2

2

sec

9 – 5tan

xdx

x∫∫∫∫

Ans. Let I =

2

2

sec

9 – 5 tan

xdx

x∫Put tan x = t

∴∴∴∴ sec2 x dx = dt

∴∴∴∴ I = 29 – 5

dt

t∫

=( ) ( )

2

3 – 5

dt

t∫

= ××××××××

1 3+ 5 1log +

1 3 3 – 5 5

tc

t

=1 3+ 5 tan

log +6 5 3 – 5 tan

xc

x

Q-9) (((( ))))

2

1

log – 9dx

x x

∫∫∫∫

Ans. Let I =( )

2

1

log – 9dx

x x

∫Put log x = t

∴∴∴∴1

xdx = dt

∴∴∴∴ I = 2 – 9

dt

t∫

= ( ) ( )2 2– 3

dt

t∫

=××××

1 – 3log +

1 3 +3

tc

t

=1 log – 3log +

6 log + 3

xc

x

=2

log

1 –

dt

a

t∫

= ( )–11sin +

logt c

x

= ( )–1 x1sin +

loga c

a



Q-13)24 5x dx−−−−∫∫∫∫

Ans. I =24 5x dx−∫

=2 5

2 –4

x dx∫

=

2

2 52 –

4x dx

∫

=

2

2

2–

2

2

5

25–

2 2 22

5log

2

xx

c

x x

+ +

+ +

=2 25

4 – 5 – log 2 4 – 52 4

xx x x c+ +

Q-14)2 4 5x x dx+ ++ ++ ++ +∫∫∫∫

Ans. I =2 +4 +5x x dx∫

=2 +4 +4+1x x dx∫

= ( )2

+2 +1x dx∫

= ( )2

2

+ 2+ 2 +1

2

1+ log + 2+ + 4 +5 +2

xx

x x x c

I =2

2

+ 2+ 4 +5

2

1+ log + 2+ + 4 +5 +2

xx x

x x x c

Q-15)29 6 7x x dx+ ++ ++ ++ +∫∫∫∫

Ans. I =29 +6 +7x x dx∫

=2 6 7

3 + +9 9

x x dx∫

=2 2 1 2

3 + + +3 9 3

x x dx∫

=

221 2

3 + +3 3

x dx ∫

=

( )

22

2

1+

1 223 + +

2 3 3

2 1 2 7+ log + + + + +3 3 3 3 9

x

x

xx x c

I =2

2

3 +1 2 7+ +

2 3 9

1 2 7+ log + + + + +

3 3 9

x xx

xx x c

=( )

( )

2

2

cos

3 – 2 1 – 2sin

sin+

3 – 2 2cos –1

xdx

x

xdx

x

∫

∫

= 2 2

cos sin+

1+ 4sin 5 – 4cos

x xdx dx

x x∫ ∫Put sin x = t

1, in first integral

∴∴∴∴ cos x dx = dt1 and

Put cos x = t2 in second integral

∴∴∴∴ sin x dx = –dt2

∴∴∴∴ I =21

2 21 2

–5 + 4 5 – 4

dtdt

t t∫ ∫

= ( ) ( ) ( ) ( )

21

2 2 2 21 2

–1 + 2 5 – 2

dtdt

t t∫ ∫

=( )–1

1

2

2

××××

tan 2

2

5 + 21 1– log +

22 5 5 – 2

t

tc

t

= ( )–11tan 2sin

2

1 5 + 2cos– log +4 5 5 – 2cos

x

xc

x



GROUP (F) – CLASS WORK PROBLEMS

Q-1) 2

5 –1

2 +3 + 7

xdx

x x∫∫∫∫

Ans. Let I = 2

–5 1

2 +3 +7

xdx

x x∫d

dx (2x2 + 3x + 7) = 4x + 3

Let 5x – 1 = A(4x + 3) + B. Then 4A = 5

and 3A + B = – 1

∴∴∴∴ A = 5

2and

53

4

+ B = –1

∴∴∴∴ B = 19

–4

∴∴∴∴ I =( )

2

5 194 + 3 –

4 4

2 + 3 + 7

x

dxx x

∫

= 2 2

5 4 +3 19–

4 42 +3 +7 2 +3 +7

x dxdx

x x x x∫ ∫= 2

2

5log 2 +3 +7

4

19 1–

3 74 2+ +2 2

x x

dx

x x

×

∫

... (i)

Now,

=2 3 7+ +2 2

dx

x x∫

=2 3 9 7 9+ + + –2 16 2 16

dx

x x

∫

=22

3 47+ +4 4

dx

x x

∫

= –1

3+

1 4tan

47 47

4 4

x

=–14 4 + 3

tan47 47

x

.... (ii)

from (i) and (ii)

I = 2 –15 19 4 +3

log 2 + 3 + 7 – tan4 47 47

xx x

(from i and ii)

Q-2) 2

2 5

+ 4 +5

xdx

x x

++++

∫∫∫∫

Ans. Let I = 2

2 +5

+ 4 +5

xdx

x x∫

=2 + 4 1

++ 4 +5 + 4 +5

xdx

x x x x

∫ 2 2

= 2 2

2 + 4 1+

+ 4 +5 + 4 +5

xdx

x x x x∫ ∫

Q-16)25 – 4 – 3x x x

e e e dx∫∫∫∫

Ans. I =2x x x5 – 4 – 3e e e dx∫

Put ex = t

ex dx = dt

I =25 – 4 – 3t t dt∫

=2 4 3

5 – –5 5

tt dt ∫

=2 2 4 19

5 – 2 + –5 25 25

t t dt ∫

=

( )

2

22

22

19

5– 2 5 2 19– – –

2 5 5 25 +

2 2 19log – + – –

5 5 5

tt

c

t t

=

2

2

x x

x

x

x x

5 – 2 4 3– –

10 5 55 +

19 2 4 3– log – – –

5 5 550

e ee

ce

e e

+



Q-3) (((( )))) 23 – 2 + +1x x x dx∫∫∫∫Ans. Let I = ( )3 – 2 + +1x x x dx∫ 2 ... (i)

3x – 2 = ( )2. + +1 +d

A x x Bdx

= A(2x + 1) + B ... (ii)

∴∴∴∴ 3x – 2 = (2A)x + (A + B)

comparing coefficients of x and constant term

on both sides,

2A = 3 and A + B = – 2

∴∴∴∴ A = 3

2 and B =

7–2

...(iii)

∴∴∴∴ From (i), (ii) and (iii)

I = ( )3 72 +1 – + +1

2 2x x x dx

∫ 2

= ( )3

2 +1 + +12

7– + +12

x x x dx

x x dx

∫∫

2

2

.....(iv)

Let I1= ( )2 +1 + +1x x x dx∫ 2

Put x2 + x + 1 = t

∴∴∴∴ (2x + 1) dx = dt

I1= .∫ t dx

=

1

2t dt∫

=

3

2+

3

2

tc

∴∴∴∴ I1= ( )

32 2

1

2+ +1 +

3x x c

Let I2=

2 + +1x x dx∫

=2 1 3+ + +

4 4x x dx∫

=

221 3

+ +2 2

x ∫

= 2

2

2

1 1+ + +1

2 2

3

2 1+ log + + + +1

2 2

x x x

x x x

∴∴∴∴ I1= ( ) ( )

32 22

2

2 7+ +1 – 2 +1 + +1

3 8

21 1– log + + + +1 +16 2

x x x x x

x x x c

=

( ) ( )

2

2 2

12 + 4 +5 +

+2 + 2x x dx

x∫

∵∵∵∵ =( )

( )

f xdx

f x

′

∫= ( )2 +f x c

=

( ) ( ) ( )

2

2 2

2 + 4 +5

+ log + 2 + + 2 + 2 +

x x

x x c

=

( )

2

2

2 + 4 +5

+ log + 2 + + 4 +5 +

x x

x x x c

Q-4)+1

9 –

xdx

x∫∫∫∫Ans. I =

+1

9 –

xdxx∫

= ××××+1 +1

9 – –1

x xdx

x x∫ ∫

= ( ) ( )

+1

9 – +1

xdx

x x∫

= 2

+1

8 – + 9

xdx

x x∫Let x + 1 = A

d

dx (8x – x2 + 9)+ B

= A (8 – 2x) + B ..... (ii)

= (8A + B) – 2Ax.

∴∴∴∴ 8A + B = 1, – 2A = 1



Q-3) 2

+3

+2 +2

xdx

x x∫∫∫∫

Ans. Let I = 2

+ 3

+ 2 + 2

xdx

x x∫ ...(i)

Let x + 3

= A d

dx (x2 + 2x + 2)+ B

= A (2x + 2) + B ...(ii)

= 2Ax + (2A + B)

GROUP (F) – HOME WORK PROBLEMS

Q-1) 2

+ 4

+5

xdx

x∫∫∫∫

Ans. Let I = 2

+ 4

+5

xdx

x∫= 2 2

1 1+4

+5 +5dx dx

x x∫ ∫=

( ) ( )2 2

2

1 2 1+ 4

2 +5 + 5

xdx

x x∫ ∫

=2 –11 4

log +5 + tan +2 5 5

xx c

Q-2) 2

5 –1

3 + +2

xdx


2

5 –1

3 + +2

xdx

x x∫ ...(i)

Let 5x + 1

= A d

dx (3x2 + x + 2)+ B

= A (6x – 1) + B ...(ii)

= 6Ax (A + B)

∴∴∴∴ 6A = 5

∴∴∴∴ A = 5

6 and

∴∴∴∴ A + B = –1

∴∴∴∴ B = –1 – A

∴∴∴∴ B = –1 – 5

6 =

11–6

...(iii)

∴∴∴∴ From (i) and (iii),

I =( )

2

5 116 +1 –

6 6

3 + + 2

xdx

x x∫= 2 2

5 6 +1 11–

6 63 + +2 3 + +2

x dxdx

x x x x∫ ∫= 2

2××××

5 11log 3 + + 2 –

26 6 3+ +3 3

dxx x

xx∫

= 2

2

5log 3 + + 2

6

11–

1 1 218+ + – +3 36 36 3

x x

dx

xx∫

= 2

22

5log 3 + +2

6

11–18 1 23

+ +6 6

x x

dx

x

∫

=2

–1

5log 3 + + 2

6

11 6 +1– tan +3 23 23

x x

xc

∴∴∴∴ A = –1

2

∴∴∴∴ B = 1 – 8A = 1–(–4) = 5

∴∴∴∴ x + 1 –1

2 (8 – 2x) + 5 ...(ii)

From (i) and (ii),

I =( )

2

18 – 2 +5

2

8 – + 9

x

dxx x∫

=2

2

1 8 – 2–2 8 – + 9

+58 – + 9

xdx

x x

dx

x x

∫∫

= ( )

( )

2

2

1– 2 8 – + 92

+525 – – 8 +16

x x

dx

x x

∫∫

=( ) ( )

2

2 2– 8 – + 9 +5

5 – – 4

dxx x

x∫

=2 –1 – 4

– 8 – +9 +5sin +5

xx x c



Q-4)+1

+3

xdx


+1

+ 3

xdx

x∫ =+1 +1

.+ 3 +1

x xdx

x x∫=

( ) ( )

+1

+3 +1

xdx

x x∫

I = 2

+1

+ 4 + 3

xdx

x x∫ ...(i)

Let

x + 1 = A d

dx (x2 + 4x + 3)+ B

= A (2x + 4) + B ...(ii)

= 2Ax + (4A + B)

∴∴∴∴ 2A = 1 ⇒⇒⇒⇒ A = 1

2

∴∴∴∴ 4A + B = 1 ⇒⇒⇒⇒ B = 1 – 4A = 1 – 2 = –1

∴∴∴∴ From (i) and (ii)

∴∴∴∴ I =( )

2

12 + 4 –1

2

+ 4 +3

x

dxx x∫

∴∴∴∴ I = 2 2

1 2 + 4–

2 + 4 + 3 + 4 + 3

x dxdx

x x x x∫ ∫

= ( )( ) ( )

2

2 2

12 + 4 +3 +

2 +1 – 1

dxx x

x∫

=

( ) ( )

2

2 2

+ 4 +3

+ log + 2+ +1 + 1 +

x x

x x c

= 2

2

+ 4 + 3

– log + 2+ + 4 + 3 +

x x

x x x c

Q-5)+ 4

3 –

xdxx∫∫∫∫

Ans. Let I =+ 4

3 –

xdxx∫ = ××××

+ 4 + 4

3 – + 4

x xdx

x x∫=

( ) ( )

+ 4

3 – + 4

xdx

x x∫

= 2

+ 4

– +12

xdx

x x∫ ...(i)

Let

x + 4 = A d

dx (x + x2 + 12)+ B

= A (1 – 2x) + B ...(ii)

= (A + B) – 2Ax

∴∴∴∴ A + B = 4 and – 2A = 1

∴∴∴∴ A = 1

–2

∴∴∴∴ B = 4 – A = 4–1

–2

= 9

2


∴∴∴∴ I =( )

2

1 9– 1 – 2 +2 2

+ +12

x

dxx x∫

=1 1 – 2

–2 – +12

xdx

x x∫ 2

9+2 – +12

dxdx

x x∫ 2

= ( )2

2 2

1 92 – +12 +

2 2 7 1– –

2 2

dxx x

x

∫

=2 –1

1–

9 2– +12 + sin +72

2

x

x x c

=2 –19 2 –1

– +12 + sin +2 7

xx x c

∴∴∴∴ 2A = 1 ⇒⇒⇒⇒ A = 1

2

∴∴∴∴ 2A + B = 3 ⇒⇒⇒⇒ B = 3 – 2A = 3 – 1

∴∴∴∴ B = 2 ...(iii)

∴∴∴∴ From i), ii) and iii)

∴∴∴∴ I =( )

2

12 +2 +2

2

+2 +2

x

dxx x∫

=2 2

1 2 + 2+ 2

2 +2 + 2 + 2 + 2

x dxdx

x x x x∫ ∫= ( )

( ) ( )

2

2 2

12 +2 + 2 +2

2 +1 + 1

dxx x

x∫

=

( ) ( ) ( )

2

2 2

+2 +2

+2log +1 + +1 + 1 +

x x

x x c

=

( )

2

2

+ 2 + 2

+ 2log +1 + + 2 + 2 +

x x

x x x c



GROUP (G) – CLASS WORK PROBLEMS

Q-1) 2

1

4 +3sindx

x∫∫∫∫

Ans. Let I = 2

1

4+ 3sindx

x∫

=( ) ( )

22

1

2 3 sin

dx

x∫

=–11 3sin 1

tan × +2 2 3

xc

=–11 3 sin

tan +22 3

xc

Q-2)cos

cos3

xdx

x∫∫∫∫

Ans. Let I =cos

cos3

xdx

x∫

= 3

cos

4cos – 3cos

x dx

x x∫= 24cos – 3

dx

x∫Divding both num. and deno. by cos2 x,

we get,

I =

2

2

sec

4 – 3sec

x dx

x∫

= ( )

2

2

sec

4 – 3 1+ tan

x dx

x∫

=

2

2

sec

1 – 3 tan

x dx

x∫Put tan x = t. Then sec2 x dx = dt

∴∴∴∴ I = 21 – 3

dt

t∫=

2

2

1

3 1–

3

dt

t

∫

=

1+

1 1 3. log +

3 1 12 –

3 3

t

c

t

=1 1+ 3 .tan

log +2 3 1 – 3 .tan

xc

x

Q-3) 2 2

1

2sin – 3cos + 7dx

x x∫∫∫∫Ans. I = 2 2

1

2sin – 3cos +7dx

x x∫Dividing Nr and Dr by cos2x

∴∴∴∴ I =

2

2 2

sec

2tan – 3+ 7sec

xdx

x x∫

= ( )

2

2 2

sec

2tan – 3 + 7 1+ tan

xdx

x x∫

=

2

2

sec

9tan + 4

x dx

x∫put tan x = t


∴∴∴∴ I = 29 + 4

dt

t∫ =

( ) ( )2 2

3 + 2

dt

t∫

=–1

××××1 3 1tan +

2 2 3

tc

=–11 3tan

tan +6 2

xc

Q-4) 2 2

1

sin + tandx

x x∫∫∫∫

Ans. I = 2 2sin + tan

dx

x x∫

=

2

2 2 2

sec

tan + tan sec

x dx

x x x∫

= ( )

2

2 2 2

sec

tan + tan 1+ tan

x dx

x x x∫

=

2

4 2

sec

tan + 2tan

x dx

x x∫Put tan x = t




I = 4 2+2

dt

t t∫

= ( )2 2 +2

dt

t t∫

= ( )

2 2

2 2

1 +2 –

2 + 2

t tdt

t t∫

= 2 2

1 1–

2 + 2

tdt dt

t t

∫ ∫

= 2 22

1 1 1–

2 2dt dt

t t

∫ ∫ +

=–11 1 1

– – tan +2 2 2

tc

t

=–11 1 1 tan

– – tan +2 tan 2 2

xc

x

GROUP (G) – HOME WORK PROBLEMS

Q-1) 2

1

4 +5sindx

x∫∫∫∫

Ans. Let I = 2

1

4+5sindx

x∫Dividing Nr and Dr by cos2x

∴∴∴∴ I =

2

2 2

sec

4sec + 5 tan

xdx

x x∫

= ( )

2

2 2

sec

4 tan +1 +5 tan

xdx

x x∫

=

2

2

sec

4 + 9tan

xdx

x∫put tan x = t


∴∴∴∴ I = 24+9

dt

t∫ =

( ) ( )2 2

2 + 3

dt

t∫ = –1

××××1 3 1tan +

2 2 3

tc

=–11 3tan

tan +6 2

xc

Q-3) 2 2

1

sin +2cos +3dx

x x∫∫∫∫

Ans. Let I = 2 2

1

sin +2cos + 3dx

x x∫Dividing Nr and Dr by cos2x

∴∴∴∴ I =

2

2 2

sec

tan + 2+ 3sec

xdx

x x∫

= ( )

2

2 2

sec

tan + 2+ 3 1+ tan

xdx

x x∫

=

2

2

sec

5 + 4tan

xdx

x∫put tan x = t


Q-2) 2

1

9 16cosdx

x++++∫∫∫∫

Ans. I = 2

1

9 16cosdx

x+∫Dividing numerator and denominator

by cos2x

I =

2

2

sec

9sec +16

xdx

x∫

= ( )

2

2

sec

9 1+ tan +16

xdx

x∫

I =

2

2

sec

9tan +25

xdx

x∫Put tan x = t


I = 2

1

9 + 25dt

t∫

= 2

2

1 1

9 5+

3

dt

t

∫

= ( )–11 1

tan +9 5 3 5 3

tc

=–11 3

tan +15 5

tc

=–11 3tan

tan +15 5

xc



Q-4) 2 2

1

4cos + sindx

x x∫∫∫∫Ans. Let I = 2 2

1

4cos + sindx

x x∫Dividing Nr and Dr by cos x

∴∴∴∴ I =

2

2

sec

4 + tan

xdx

x∫put tan x = t


∴∴∴∴ I = 24 +

dt

t∫ =( ) ( )

2 22 +

dt

t∫

=–11

tan +2 2

tc

=–11 tan

tan +2 2

xc

∴∴∴∴ I = 25 + 4

dt

t∫

= ( ) ( )2 2

5 + 2

dt

t∫

=–1

××××1 2 1

tan +25 5

tc

=–11 2tan

tan +2 5 5

xc

Q-5) 2 2

1

9cos 16sindx

x x++++∫∫∫∫

Ans. I = 2 2

1

9cos +16sindx

x x∫Dividing numerator and denominator

by cos2x

I =

2

2

sec

9 +16tan

xdx

x∫Put tan x = t


I = 2

1

16 + 9dt

t∫

I = 2

2

1 1

16 3+

4

dt

t

∫

=( )

–11 1tan +

16 3 4 3 4

tc

=–11 4

tan +12 3

tc

I =–11 4tan

tan +12 3

xc

GROUP (H) – CLASS WORK PROBLEMS

Q-1)1

5+ 4cos x∫∫∫∫

Ans. Let I =5 + 4cos

dt

x∫Put tan

2

x= t,

∴∴∴∴ dx = 2

2

1+

dt

t and cos x =

2

2

1 –

1+

t

t

I =

2

2

2

2

1+

1 –5 – 4

1+

dt

t

t

t

∫

I = ( ) ( )2 2

2

5 1+ – 4 1 –

dt

t t∫

∴∴∴∴ I = 2

2

+ 9

dt

t∫ =( ) ( )

2 22

+ 3

dt

t∫

=–11

2 tan +3 3

tc

=–1

tan2 2tan +3 3

x

c

Q-2)1

3 +3sin 2cosdx

x x++++∫∫∫∫

Ans. Let I = 1

3 + 3sin + 2cosdx

x x∫Put tan =

2

xt

∴∴∴∴ sin x = 2

2

1+

t

t, , , , cos x =

2

2

1 –

1+

t

t

∴∴∴∴ I =

2

2

2 2

2

1+

2 1 –3 + 3 + 2

1+ 1+

dt

t

t t

t t

∫



Q-4)1+ cos cos

cos + cos

xdx

x

αααα

αααα∫∫∫∫

Ans. Let I =αααα

αααα

1+ cos cos

cos + cos

xdx

x∫Put tan =

2

xt ⇒⇒⇒⇒ 2

2=1+

dtdx

t &

2

2

1–cos =

1+

tx

t

I =

2 2α α αα α αα α αα α α

αααα

sin + cos + cos .cos

cos + cos

xdx

x∫

I =( )2 α α αα α αα α αα α α

αααα

sin + cos cos + cos

cos + cos

xdx

x∫

I =2 α αα αα αα α

ααααsin + cos 1.

cos + cos

dxdx

x∫ ∫ ... (i)

Let I1=

ααααcos + cos

dx

x∫ from above sum

I1=

αααα

αααααααα

cot tan1 2 2log

sincot – tan

2 2

x

x

+

∴∴∴∴ I =2

∝∝∝∝

∝∝∝∝∝∝∝∝∝∝∝∝

∝∝∝∝

cot + tan1 2 2sin . log

sincot – tan

2 2

+ .cos +

x

x

x c

∴∴∴∴ I =

∝∝∝∝

∝ ∝∝ ∝∝ ∝∝ ∝∝∝∝∝

cot + tan2 2sin .log + .cos +

cot – tan2 2

x

x cx

Q-5) I =1

3cos2 5++++∫∫∫∫ dxx

Ans. Let I = 1

3cos2 +5dx

x∫Put tanx = t

∴∴∴∴ cos 2x =

2

2

1 –

1+

t

t

Q-3)cos + cos

dx

xαααα∫∫∫∫

Ans. Let I =ααααcos + cos

dx

x∫Put tan =

2

xt ⇒⇒⇒⇒ 2

2=1+

dtdx

t &

2

2

1–cos =

1+

tx

t

I =

( )2

2

2αααα ××××

2

1 –cos + 1+

1+

dt

tt

t

∫

I = ( ) ( )2 2αααα

2

cos 1+ + 1 –

dt

t t∫

I = 2 2α αα αα αα α

2

cos + cos . +1 –

dt

t t∫

I = ( ) ( )2α αα αα αα α

2

1+ cos – 1 – cos

dt

t∫

I =2 2 2α αα αα αα α

2

2cos – 2sin2 2

dt

t

∫

I =2 2 2α αα αα αα α

2

2sin cot –2 2

dt

t

∫

I =2 α αα αα αα α

1 1. .

sin 2cot2 2

αααα

αααα

cot +2

log +

cot –2

t

c

t

I =

αααα

αααααααα

cot + tan1 2 2

log +sin

cot – tan2 2

x

cx

∴∴∴∴ I =

∝∝∝∝

∝∝∝∝∝∝∝∝

cot + tan2 2

cos log +

cot – tan2 2

x

ec cx

=( ) ( )2 2

2

3 1+ + 6 +2 1 –

dt

t t t∫

=( ) ( )

2 22

+ 3 – 2

dt

t∫

=××××

1 +3 – 22 log +

2 2 +3+2

tc

t

=

tan +11 2log +

2tan +5

2

x

cx



Q-6) I =1

2sin2 1++++∫∫∫∫ dxx

Put tan x = t

x = tan–1t 2

1

1dx dt

t

=+

∴∴∴∴ sin 2x =2

2

1

t

t+

∴∴∴∴ I =( )2

2

1 1

2 12 11

dtt t

t

++ +

∫

= 2

1

4 1dt

t t+ +∫

I =( )

2–3

1

2dt

t +∫

=

( ) ( )22

1

2 – 3

dt

t +∫

=1 2 – 3

log2 3 2 3

tc

t

++

+ +

=1 tan 2 – 3

log2 3 tan 2 3

xc

x

++

+ +

GROUP (H) – HOME WORK PROBLEMS

Q-1) 1

3 + 2sinx∫∫∫∫

Ans. I = 1

3+ 2sinx∫Put tan

2

x

= t, then 21sec =1

2 2

x dx

dt

∴∴∴∴ dx =2

2

sec2

dt

x

=2

2

1+ tan2

dt

x

= 2

2

1+

dt

t

∴∴∴∴ I = 2

2

1 2.

2 1+3 + 2

1+

t

t t

t

∫

∴∴∴∴ I = ( )

2

22

1+.1+3 1+ + 4

t dt

tt t∫

∴∴∴∴ I = 2

12

3 + 4 + 3dt

t t∫=

2

2 1

43+ +13

dt

t t∫

∴∴∴∴ I = 2

2 1

3 2 4+ – +13 9

dt

t

∫

∴∴∴∴ I = 2

2 1

3 2 5+ –3 9

dt

t

∫

∴∴∴∴ I = –1

2+

2 1 3tan +3 55

33

t

c

∴∴∴∴ I =–12 3 +2

tan +5 5

tc

∴∴∴∴ I =–1

3tan +22 2tan +

5 5

x

c

dx = 21+

dt

t

∴∴∴∴ I = 22

2

1

1+1 –3 +51+

dt

tt

t

∫

∴∴∴∴ I = ( ) ( )2 23 1 – +5 1+

dt

t t∫

= 22 +8

dt

t∫=

2

1

2 + 4

dt

t∫=

( ) ( )2 2

1

2 + 2

dt

t∫

=–1

××××

1tan +

2 2 2

tc

= –11 tantan +

4 2

xc



Q-4) 1

1+ cos cos xαααα∫∫∫∫

Ans. I = αααα

1

1+ cos cos x∫Put tan

2

x

= t, then 21

sec =12 2

x dx

dt

∴∴∴∴ dx =2

2

sec2

dt

x

=2

2

1+ tan2

dt

x

= 2

2

1+

dt

t

∴∴∴∴ I =22

2αααα

1 2.1+1 –

1+ cos1+

dt

tt

t

∫

∴∴∴∴ I =

2

22 2αααα

1+2 .

1+1+ + cos 1+

t dt

tt t ∫

∴∴∴∴ I =2 2 2α αα αα αα α

12

2sin + 2cos2 2

dt

t

∫

Q-3)1

2+ sin + cosdx

x x∫∫∫∫

Ans. Let I = 1

2+sin + cosdx

x x∫Put tan =

2

xt

∴∴∴∴ sin x = 2

2

1+

t

t, , , , cos x =

2

2

1 –

1+

t

t

Q-2) 1

4 – 5cos x∫∫∫∫

Ans. I = 1

4 – 5cosdxx∫

Put tan 2

x

= t, then 21

sec =12 2

x dx

dt

∴∴∴∴ dx =2

2

sec2

dt

x

=2

2

1+ tan2

dt

x

= 2

2

1+

dt

t

∴∴∴∴ I =( )

1 2.

1 – 1+4 – 5

1+

t

t t

t

∫ 2 2

2

∴∴∴∴ I =( ) ( ) ( )

1+

1+4 1+ – 5 1+

t dt

tt t ∫

2

22 2

∴∴∴∴ I = 2

12

9 –1dt

t∫∴∴∴∴ I =

2

2 1

19–

9

dt

t

∫

∴∴∴∴ I =( )

1–

2 1 3log +119

+233

t

c

t

∴∴∴∴ I =

1tan –

1 2 3log +

13tan +

2 3

x

cx

∴∴∴∴ I =

3tan –11 2log +

33tan +1

2

x

cx

dx = 2

2

1+

dt

t

∴∴∴∴ I = 2 2

2 2

1 2

2 1 – 1+2+ +

1+ 1+

dt

t t t

t t

∫

=( ) ( )2 2

2

2 1+ +2 + 1 –

dt

t t t∫

= 2 + 2 + 3

dt

t t∫

=( ) ( )

22+1 + 2

dt

t∫

=–11 +1

tan +2 2

tc

= –1

+11 2tan +

2 2

xt

c



∴∴∴∴ I =1 1

sin + cos2 2

dt

t

∫2 2 2α αα αα αα α

∴∴∴∴ I =2 2 2α αα αα αα α

1 1

sin + cot2 2

dt

t

∫

∴∴∴∴ I = –1

2 α α αα α αα α αα α α

1 1 1tan

sin cos cos2 2 2

∴∴∴∴ I =–1 αααα

α αα αα αα α

1 1tan tan

2sin cos

2 2

t

∴∴∴∴ I = 2cosec (αααα) tan–1αααα

tan tan +2 2

xc

Q-5)1

5+ 4 sindxx∫∫∫∫

Ans. Let I = 1

5 + 4sindxx∫

Put tan =2

xt

∴∴∴∴ sin x = 2

2

1+

t

t,,,, dx = 2

2

1+

t

t

∴∴∴∴ I =2

2

××××1 2

2 1+5 + 4

1+

dt

t t

t

∫

= ( )2

2

5 1+ + 8

dt

t t∫

= 22

5 +8 +5

dt

t t∫=

2

2

85+ +15

dt

t t∫

=2

2

8 64 645+ + +1 –5 100 100

dt

t t∫

= 2 2

2

5 8 6+ +10 10

dt

t

∫

=–1

××××

8+

2 1 10tan +665

1010

t

c

=–12 5 + 4

tan +5 3

tc

∴∴∴∴ I =–1

5tan + 42 2tan +5 3

x

c

Q-6)1

2 – 3sin2dx

x∫∫∫∫

Ans. Let I =1

2 – 3sin2dx

x∫Put t = tan x tan–1(t) = x

∴∴∴∴ dx = 2

1

1+dt

t

sin 2x = 2

2tan

1+ tan

x

x= 2

2t

1+ t

I =( )2

2

1

1+

22 – 3

1+

dtt

t

t

∫ = ( ) ( )22 1+ – 3 2

dt

t t∫

= 2

1

2 – 6 + 2dt

t t∫ = 2

1 1

2 – 3 +1dt

t t∫=

2

1 1

9 92– 3 + – +1

4 4

dt

t t∫

= 22

1 1

2 3 5– –2 2

t

∫

= ××××

3 5– –

1 1 2 2log +2 3 55

– +22 22

t

c

t

=1 2 – 3 – 5

log +2 5 2 – +3 + 5

tc

t

=( )

( )

2tan – 3 + 51log +

2 5 2tan – 3 – 5

x

c

x



Q-7)1

2 – 3cos2∫∫∫∫ dxx

Ans.Let I =1

2 – 3cos2dx

x∫Put tan = t

∴∴∴∴ dx = 2

1

1+dt

t,

cos 2x =

2

2

1 –

1+

t

t

I =22

2

1

1+1 –2 – 3

1+

dt

tt

t

∫

= ( ) ( )2 22 1+ – 3 1 –

dt

t t∫

= 25 –1

dt

t∫

=( ) ( )

2 25 – 1

dt

t∫

= ××××1 5 –1 1log +

2 5 +1 5c

t

=1 5 tan –1

log +2 5 5 tan +1

xc

x

Q-8)1

3 2sin2 4cos2+ ++ ++ ++ +∫∫∫∫ dxx x

Ans. I =1

3+ 2sin2 + 4cos2dx

x x∫Put tan x = t x = tan–1 t

∴∴∴∴ sin x = 2

2

1+

t

t

cos x =

2

2

1 –

1

t

t+

dx = 21+

dt

t

I =( )2 2

2 2

1

2 1 – 1+3 +2 + 4

1+ 1+

dt

t t t

t t

×

∫

= 2 2

1

3+ 3 + 4 + 4 – 4dt

t t t∫

Q-9)1

3sin 4 cos 5+ ++ ++ ++ +∫∫∫∫ dxx x

Ans. I =1

3sin 4cos 5+ +∫ dxx x

Put tan 2

x = t

∴∴∴∴ x = 2tan–1 t

dx = 2

2

1+dt

t

∴∴∴∴ sin x = 2

2

1+

t

t,

cos x =

2

2

1 –

1+

t

t

I = 22

2 2

1 2

1+2 1 –3 + 4 +51+ 1+

dt

tt t

t t

∫

= 2 2

12

6 + 4 – 4 +5 +5dt

t t t∫

= 2 2

12

6 + 4 – 4 +5 +5dt

t t t∫= 2

12

+ 6 – 9dt

t t∫= ( )

–22 +3t dt∫ =

–12 +

+3c

t

I =–2

+

tan + 32

cx

= 2

1

7 + 4 –dt

t t∫= 2

1

11 – 4 – 4 –dt

t t∫

=( ) ( )

2 2

1

11 – – 2dt

t∫

=( )

( )

11+ – 22log +

2 11 11 – – 2

tc

t

=2 11 + tan – 2

log +2 11 11 – tan +2

xc

x

I =1 tan + 11 – 2

log +2 11 11 +2 – tan

xc

x



Q-10)1

3 4 sin++++∫∫∫∫ dxx

Ans. I =1

3 4sin+∫ dxx

Put tan 2

x = t

∴∴∴∴ x = 2tan–1 t

∴∴∴∴ sin x = 2

2

1+dt

t

dx = 2

2

1+

t

tdt

I =( )2

2

1 2

2 1+3+ 41+

dt

t t

t

∫

= ( )2

12

3 1+ +8dt

t t∫

= 2

12

3+3 +8dt

t t∫=

2

2 1

83+ +13

dt

t t∫

=2

2 1

4 16 73+ 2 + –

3 9 9

dt

t t

∫

=2

2 1

3 4 7+ –3 9

dt

t

∫

= 22

2 1

3 4 7+ –3 3

dt

t

∫

=

4 7–

2 1 3 3log3 4 772

3 33

t

t

+

+ +

=

3tan 4 – 71 2log7 3tan 4 7

2

+

+

+ +

x

cx

GROUP (I) – CLASS WORK PROBLEMS

Q-1) 1

3 sin + 4cosdx

x x∫∫∫∫

Ans. Let I = 1

3sin + 4cosdx

x x∫

=1 1

3 45sin + cos

5 5

dx

x x∫

=θ θθ θθ θθ θ

1

45cos sin + sin cos

5

dx

x x∫

Where cos θθθθ =3

5 and

sin θθθθ =4

5

∴∴∴∴ tan θθθθ =4

3

∴∴∴∴ I =( )θθθθ

1

5 sin +

dx

x∫

∴∴∴∴ θθθθ = tan–14

3

= ( )θθθθ1

cos +5

ec x dx∫

=θθθθ1

log tan + +5 2 2

xc

=

–1 4+ tan

1 3log tan +

5 2

x

c

Q-2)1

cos + 3 sindx

x x∫∫∫∫

Ans. Let I =1

1.cos + 3 sindx

x x∫Let 3 = r cos αααα and 1 = r sin αααα

∴∴∴∴2

3 + 12 = r2 and tan α α α α = 1

3

∴∴∴∴ 2 = r and α α α α = ππππ

6

3 sinx + 1 cosx = r cosαααα sinx + r sin αααα cosx



Q-3)1

1+ sindx

x∫∫∫∫

Ans. Let I =1

1+sindx

x∫=

1

cos + sin2 2

dxx x∫

=1

1 12 cos + sin2 22 2

dx

x x∫

=π ππ ππ ππ π

1

2 sin .cos + cos sin4 2 4 2

dx

x x∫

=ππππ

1

2sin +

2 4

dx

x

∫

=ππππ1

cos +2 42

xec dx

∫

=

ππππ

××××

+1 12 4log tan +

122

2

x

c

=ππππ1

log tan + +4 82

xc

GROUP (I) – HOME WORK PROBLEMS

Q-1) 1

2sin – cosdx

x x∫∫∫∫

Ans. I = 1

2sin – cosdx

x x∫2 = r cos ∝∝∝∝ and 1 = r sin ∝∝∝∝

∴∴∴∴ (2)2 + (1)2 + r2 and tan ∝ ∝ ∝ ∝ = 2

∴∴∴∴ r = 5 , ∝ ∝ ∝ ∝ = tan–1(2)

∴∴∴∴ I =2sin –1.cos

dx

x x

=∝ ∝∝ ∝∝ ∝∝ ∝5 cos .sin – 5 sin .cos

dx

x x∫= ( )∝∝∝∝5 sin –

dx

x∫

= ( )∝∝∝∝1

cos –5

ec x dx∫=

∝∝∝∝1 –log tan +

25

xc

= ( )–1– tan 21log tan +

25

xc

Q-2)1

3sin – 4cosdx

x x∫∫∫∫

Ans. Let I =1

3sin – 4cosdx

x x∫= 1

3 45sin – cos

5 5

dx

x x∫

Let, 3

5 = cos a and

4

5 = sina

∴∴∴∴ tan a = αααα

αααα

sin 4=

cos 5

∴∴∴∴ a = tan–1 4

3

So,

I =α αα αα αα α

1 1

5 cos sin – sin cosdx

x x∫

= ( )αααα

1 1

5 sin –dx

x∫

∴∴∴∴ 3 sinx + cosx = r sin (x + αααα)

∴∴∴∴ I = ( )ααααsin +

dx

r x∫

∴∴∴∴ I = ( )αααα1

cos +ec x dxr ∫

∴∴∴∴ I =αααα1 +

log tan +2

xc

r

∴∴∴∴ I =

ππππ+

1 6log tan +2 2

x

c



= ( )αααα1

cos –5

ec x dx∫

=αααα1 –

log tan +5 2

xc

=

–1 4– tan

1 3log tan +

5 2

x

c

Q-3)1

sin + 3 cosdx

x x∫∫∫∫Ans. Let, I =

1

sin + 3 cosdx

x x∫=

1 1

2 1 3sin + cos

2 2

dx

x x∫


1 1

2cos sin + sin cos

3 3

dx

x x∫

=ππππ

1 1

2sin +

3

dx

x

∫

=ππππ1

cos +2 3

ec x dx ∫

=ππππ1

log tan + +2 2 6

xc

Q-4)1

1+ sin2dx

x∫∫∫∫Ans. Let, I =

1

1+sin2dx

x∫

=( )

2

1

cos + sin

dx

x x∫

=cos +sin

dx

x x∫

=

1

1 12 cos + sin22 2

dx

xx

∫


1

2 sin .cos + cos sin4 2 4

dx

xx

∫

=ππππ

1

2 sin +4

dx

x

∫

=ππππ1

cos +42

ec x dx ∫

=

ππππ+

1 4log tan +22

x

c

=ππππ1

log tan + +2 82

xc

Q-5)1

5cos 12sindx

x x++++∫∫∫∫Ans. I =

+∫1

5cos 12sindx

x x

Put tan2

x = t

∴∴∴∴ x = 2 tan–1t

dx = 2

2

1+

dt

t, sin x = 2

2

1+

dt

t, cos x =

2

2

1 –

1+

t

t

I =( )2 2

2 2

1 2

1 – 2 1+5 +121+ 1+

dt

t t t

t t

∫

= 2

12

5 – 5 +24dt

t t∫=

2

2 1

2451 – +

5

dtt

t∫

=2

2 1

5 169 12 144– – 2 +

25 5 25

dt

t t

∫

= 2 2

2

2 1

5 13 12– –

5 5

dt

t

∫

=

+

+

+

∫13 12

–2 1 5 5log

13 12135–2

5 55

t

c

t

I =

1+ tan

1 5 2log +13

5 – tan2

x

cx



GROUP (J) – CLASS WORK PROBLEMS

Q-1)

2

2 2–

xdx

a x∫∫∫∫

Ans. I =–

xdx

a x∫2

2 2

Put x = a sin θθθθ

∴∴∴∴ dx = a cos θ θ θ θ d θθθθ

∴∴∴∴ I =

2 2

2 2 2

θ θ θθ θ θθ θ θθ θ θ

θθθθ

sin . cos

– sin

a a d

a a∫

=2 2

θ θθ θθ θθ θsina d∫=

2θθθθ

1 – cos2

2a d∫

=

2

θ θθ θθ θθ θ1

– sin2 +2 2

ac

= [ ]2

θ θ θθ θ θθ θ θθ θ θ– sin cos +2

ac

We resubstitute for sin θ θ θ θ and cos θ θ θ θ as follows.

∵∵∵∵ sin θ θ θ θ = x

a, we draw a triangle with one angle

900 and another marked as θθθθ. The opposite

side of θθθθ is taken as x and hypotenuse as a,

giving sin θθθθ = x

a. The remaining side

is 2 2–a x .

∴∴∴∴ cos θ θ θ θ = 2 2–a x

a.

∴∴∴∴ I =

2 2 2–1 –

sin – . +a x x a x

ca a a a

x a

θ

a x2 2 –

Q-2)–

xdx

a x∫∫∫∫

Ans. Let I =–

xdx

a x∫Put x = a sin2 θθθθ

∴∴∴∴ dx = 2a sin θ θ θ θ cos θθθθ dθθθθ

∴∴∴∴ I =

2

2

θθθθ

θθθθ

sin

– sin

a

a a∫ .(2a sin θθθθ. cos θθθθ dθθθθ)

=2

θ θθ θθ θθ θ2 sina d∫

=θθθθ

θθθθ1 – cos2

22

a d ∫

=θθθθ

θθθθsin2

– +2

a c

= a(θ θ θ θ – sin θθθθ. cos θθθθ)+c

=–1 . –

sin – +x x a x

a ca a

= ( )–1sin – – +x

a x a x ca

x

a x –

θ

a

Q-3) I =–

++++∫∫∫∫a x

dxa x

Put x = a cos 2θθθθ ⇒⇒⇒⇒ cos 2θθθθ

=–11

cos2

x x

a aθ

⇒ =

∴∴∴∴ dx = – 2a sin 2θθθθ dθθθθ

∴∴∴∴ I = – cos2

– 2 sin2cos2

a aa d

a a

θθ θ

θ+∫

=1 – cos2

–2 .sin21 cos2

a dθ

θ θθ+∫

=sin

–2 .2sin coscos

a dθ

θ θ θθ∫



Q-5) 2 2

1

–1x x

Ans. Let I = 2 2

1

–1dx

x x∫Put x = sec θθθθ

∴∴∴∴ dx = sec θ θ θ θ tan θ θ θ θ dθθθθ

∴∴∴∴ I = 2 2θ θθ θθ θθ θ

1

sec sec –1∫ sec θθθθ tan θθθθ dθθθθ

= 2

θ θθ θθ θθ θθθθθ

θ θθ θθ θθ θ

sec tan

sec tand∫

=1

secd∫ θθθθ

θθθθ

= θ θθ θθ θθ θcos d∫= sinθθθθ + c .... (i)

x

1

θ

x 2 – 1

Now, sec θθθθ = x

∴∴∴∴ cos θθθθ =1

x

∴∴∴∴ sin θθθθ =–1x

x

2

....[from figure]

From (i), we get

I =–1

+x

cx

2

Q-4)(((( ))))

22 2

1

+dx

x a∫∫∫∫

Ans. Let I =( )2 2

1

+dx

x a∫Put x = a tan θθθθ

∴∴∴∴ dx = a sec2 θ θ θ θ dθθθθ

∴∴∴∴ I =( )

22 2 2

θθθθ

1

+ tan +a a∫ .(a sec2 θθθθ dθθθθ)

=

2

4 4

θ θθ θθ θθ θ

θθθθ

sec

sec

a d

a∫

=2

3θ θθ θθ θθ θ

1cos d

a ∫

=

2

3

θθθθθθθθ

1 1+ cos

2d

a

∫

=

2

3

θθθθθθθθ

1 sin+ +

22c

a

∫

= ( )3

θ θ θθ θ θθ θ θθ θ θ1

+sin .cos +2

ca ∫

x

a

θ

x a2 2 +

=2–2 2sina dθ θ∫

= ( )–2 1 – cos2a dθ θ∫=

sin2–2 –

2a

θθ

=–1 21 1

–2 cos + 1 – cos 2 +2 2

xa c

aθ

=

2–1

2– cos + 1 – +

x xa a c

a a

=–1 2 2– cos + – +x

a a x ca

I =–1

3 2 2 2 2

1tan + .

2 + +

x x a

aa x a x a

I =–1

3 2 2

1tan + +

2 +

x axc

aa x a



GROUP (J) – HOME WORK PROBLEMS

Q-1)2 2

2

–a xdx

x

Ans. I =2 2

2

–a xdx

x

Put x = a sin θθθθ

∴∴∴∴ dx = a cos θ θ θ θ dθθθθ

∴∴∴∴ I =

2 2 2

2 2

θθθθ

θθθθ

– sin

sin

a a

a∫ a cos θθθθ dθθθθ

= 2 2

θ × θ θθ × θ θθ × θ θθ × θ θ

θθθθ

cos cos

sin

a a d

a∫

=

2

2

θθθθθθθθ

θθθθ

cos

sind∫

=2

θ θθ θθ θθ θcot d∫

= ( )2θ θθ θθ θθ θcos –1ec d∫

=2θ θ θθ θ θθ θ θθ θ θcos –ec d d∫ ∫

= –cot θθθθ – θ θ θ θ + c ....(i)

a x

θ

a 2 – x

2

Now, x = a sin θθθθ

∴∴∴∴ sin θθθθ =x

a

∴∴∴∴ θ θ θ θ = sin–1x

a

Also, cos θθθθ =–a x

a

2 2

....[from figure]

∴∴∴∴ cot θθθθ =θθθθ

θθθθ

cos

sin

=

–a x

ax

a

2 2

=2 2–a x

x

From (i), we get

I =2 2

–1–– sin +

a x xc

x a

Q-2)1 –

1+

xdx

x∫∫∫∫

Ans. Let I =1 –

1+

xdxx∫

Put x = cos θθθθ

∴∴∴∴ dx = –sin θ θ θ θ dθθθθ

∴∴∴∴ I =θθθθ

θθθθ

1 – cos

1+ cos∫ (–sin θθθθ dθθθθ)

=

2

2

θθθθ


θθθθ

2sin2 –2sin .cos

2 22sin

2

d ∫

=

θθθθ


θθθθ

sin2 –2sin .cos

2 2cos

2

d ∫

=2 θθθθ

θθθθ–2 sin2d∫

=θθθθ

θθθθ1 – cos

–22

d ∫

= ( )θ θθ θθ θθ θcos –1 d∫= sin θθθθ – θ θ θ θ + c

1

x

θ

1 – x2

∴∴∴∴ I = 21 – x – cos–1(x) + c



Q-4)–a x

dxx

Ans. I =–a x

dxx

Put x = a sin2 θθθθ

∴∴∴∴ dx = 2a sin θ θ θ θ cos θ θ θ θ dθθθθ

∴∴∴∴ I =

2

2

θθθθ

θθθθ

– sin

sin

a a

a∫ 2a sin θθθθ cos θ θ θ θ dθθθθ

=

2

2

θθθθ

θθθθ

cos

sin

a

a∫ . 2a sin θθθθ cos θ θ θ θ dθθθθ

=2

θ θθ θθ θθ θ2 cosa d∫=

θθθθθθθθ

1+cos22

2a d∫

θ

ax

a x–

=θθθθ

θθθθsin2

. + +2

a c

= a(θθθθ + sin θ θ θ θ cos θθθθ) + c

=–1 . –

sin + +x x a x

a ca a

= ( )–1sin + – +x

a x a x ca

GROUP (K) – CLASS WORK PROBLEMS

Q-1) cosx x dx∫∫∫∫

Ans. Let I = .cosx x dx∫We know that,

uvdx∫ = – .du

u v dx v dx dxdx

∫ ∫

Q-3)

(((( ))))3

2 2

1

+a x∫∫∫∫

Ans. Let I =

( )3

2 2

1

+

dx

a x∫

Put x = a tan θθθθ

∴∴∴∴ dx = a sec2 θ θ θ θ dθθθθ

∴∴∴∴ I = ( )

2

32 2 2

θθθθθθθθ

θθθθ

sec

+ tan

ad

a a∫

=

( )

2

32 2

θθθθθθθθ

θθθθ

sec

1+ tan

ad

a

∫

=

2

32 2

θθθθθθθθ

θθθθ

sec

sec

ad

a ∫

=

2

6 6

θθθθθθθθ

θθθθ

sec

sec

ad

a∫ =

2

3 3

θθθθ

θθθθ

sec

sec

a

a∫

= 2

θθθθ

θθθθ

1

sec

d

a ∫ = 2θ θθ θθ θθ θ

1cos d

a ∫= 2

θθθθ1sin + c

a.... (i)

a

x

θ

x 2 – a

2

Now, x = a tan θθθθ

∴∴∴∴ tan θθθθ =x

a

∴∴∴∴ sin θθθθ = 2 2+

x

x a....[from figure]

From (i), we get

I =2 2 2

1. +

+

xc

a x a

=2 2 2

++

xc

a x a



Q-2)x

xe dx∫∫∫∫

Ans. Let I =xxe dx∫

= ( ) ( ) 1x x– +

dx e dx e dx x dx c

dx

∫ ∫

= 1x x– .1 +xe e dx c∫

= xex – ex + c

Q-3)3 logx xdx∫∫∫∫

Ans. Let I =3 logx xdx∫

∴∴∴∴ I = ( ) 3log x x dx∫

∴∴∴∴ I =

( )

3

3

log

– log

x x dx

dx x dx dx

dx

∫∫ ∫

∴∴∴∴ I =

4 41log – .

4 4

x xx dx

x

∫

∴∴∴∴ I =

43log 1

–4 4

x xx dx∫

∴∴∴∴ I =4 4log 1

– +4 4 4

x x xc

∴∴∴∴ I = ( )4

4log –1 +16

xx c

Q-4)1+ sin

xdx

x∫∫∫∫

Ans. Let I =1+sin

xdx

x∫I = ××××

1– sin

1+sin 1– sin

x xdx

x x∫I = 2

– sin

1– sin

x x xdx

x∫

I = 2

– sin

cos

x x xdx

x∫I =

2sec – sec tanx x x x dx∫uv dx∫ = –

duv dx v dx dx

dx

∫ ∫ ∫

∴∴∴∴ I = tan – 1.tan

– sec – 1.sec

x x x dx

x x x dx

∫∫

= x tan x – log |sec x|

– [x sec x – log |sec x + tan x|] + c

= x tan x – log |sec x|

+ log|sec x + tan x| + c

= x (tan x – sec x)

+ log sec + tan

+sec

x xc

x

= x (tan x – sec x) + log|1 + sin x| + c

.cosx x dx∫ =

( ) 1

. cos

– cos +

x x dx

dx dx x dx c

dx

∫∫ ∫

∴∴∴∴ x. sin x – 1sin .1 +x dx c∫∴∴∴∴ x sin x + cos x + c

Q-5)2cosx x dx∫∫∫∫

Ans. Let I =2cosx x dx∫

=2 2cos – cos .

dx x dx x dx x dx

dx

∫ ∫ ∫

... (i)

∴∴∴∴2cos x dx∫

=1+cos2

2

xdx∫

=1 sin2

+2 2

xx

and 1 sin2

+2 2

xx dx ∫



Q-6)2. xx e dx∫∫∫∫

Ans. Let I =2 x.x e dx∫

2 x.x e dx∫

= ( ) ( )2 2

1

x x– +d

x e dx e dx x dx cdx

∫ ∫

= ( )2

1

x x. – . 2 +x e e x dx c∫=

2

1

x x. – 2 . +x e x e dx c∫

∴∴∴∴ I = ( )2

2

x x x– 2 – .1 +x e x e e dx c ∫

= x2ex – 2xex + 2ex + c.

Q-7)–1. tanx x dx∫∫∫∫

Ans. Let I = ( )–1tanx x dx∫

= ( )–1tan .x x dx∫

∴∴∴∴ =

2 2

12

1tan . – . +

2 2 1+

x xx dx c

x

∫–1

=( )2

2

12

+1 –11 1tan – +

2 2 1+

xx x dx c

x∫–1

=2

12

1 1 1tan – 1 – +

2 2 1+x x dx c

x

∫–1

=2 –11 1tan – – tan + .

2 2x x x x c

–1

Q-8)–1tan x dx∫∫∫∫

Ans. Let I =–1tan x dx∫

dx = ( ) ( )–1tan . 1x dx∫

= ( )–1tan 1 – tan 1d

x dx x dx dxdx

∫ ∫ ∫–1

= ( ) ( )–1

2

1tan – .

1+x x x dx

x∫

= –1

2

1 2tan –

2 1+

xx x dx

x∫Now,

d

dx(1 + x2) = 2x and

( )

( )

'f x

f x∫ = log|f(x) + c

∴∴∴∴ I = x tan–1x – 1

2 log(1 + x2)+c

( )21+ > 0x∵

Q-9) (((( ))))2

log x dx∫∫∫∫Ans. = (log x)2 ( )

21 – log 1

ddx x dx dx

dx

∫ ∫ ∫

= log x2.x – × ×× ×× ×× ×1

2log x dxx∫

= x (log x)2 – ( )2log 1x dx∫= x (log x)2 – 2 I

1

I1= ( ) ( )log 1x dx∫= ( )log 1 – log 1

dx dx x dx dx

dx

∫ ∫ ∫

= ( ) ××××1

log –x x xdxx∫

= x log x – x + c

= x (log x – 1) + c

I = x (log x)2 – 2x (log x – 1) + c

=

21 1+ cos2

2 2 8

xx

∴∴∴∴ I =1 sin2 1 sin2. + – +2 2 2 2

x xx x x dx

∫

∴∴∴∴ I =

2 2.sin2 1+ – + cos2 +

2 4 4 8

x x x xx c

∴∴∴∴ I =

2 sin2 cos2+ + +

4 4 8

x x x xc



Q-10) (((( ))))2 2log + +∫∫∫∫ x x a dx

Ans. I = ( )2 2log + +x x a dx∫I = ( )2 21.log + +x x a dx∫= ( )

( )

2 2

2 2

log + +

1 – 1 . log + + .

x x a

ddx dx x x a dx

dx

∫ ∫ ∫

= ( )2 2

2 2 2 2

. log + +

2– 1+

+ + 2 +

x x x a

x xdx

x x a x a

∫

= ( )2 2

2 2. log + + –

+

xx x x a dx

x a∫In the second integral

Put x2 + a2 = t

∴∴∴∴ x dx =2

dt

∴∴∴∴ I = ( )2 2

2 2

1 2.log + + –

2 +

xx x x a dx

x a∫= ( ) ( )2 2 2 21

.log + + – 2 + +2

x x x a x a c

= ( )2 2 2 2. log + + – + +x x x a x a c

Q-11) cos xdx∫∫∫∫Ans. Let I = cos xdx∫

Put x = t

Then x = t2 and dx = 2t dt

I = ( )cos .2t t dt∫I = 2 cost tdt∫I = ( )2 cos – cos

dt tdt t tdt dt

dt

∫ ∫ ∫

I = 2 sin – sint t tdt ∫

I = 2(t sin t + cos t) + c

I = 2t sin t + 2 cos t + c

I = 2 sin + 2cos +x x x c

Q-12) (((( ))))sin . log cosx x dx∫∫∫∫Ans. Let I = ( )sin .log cosx x dx∫

Put cos x = t ⇒⇒⇒⇒ sin x dx = – dt

I = ( )log cos sinx x dx∫I = ( )log t dt−∫I = ( )– log .t dt∫I = ( ) ( )– log 1. – log 1

dt dt t dt dt

dt

∫ ∫ ∫

I = ( ) ( )1

– log –t t t dtt

∫I = – log – 1t t dt

∫

I = –t log t + t + c

I = t (1 – log t) + c

I = cos x {1 – log (cos x)} + c

Q-13)23 x

x e dx∫∫∫∫Ans. I =

23 xx e dx∫=

22 x.x x e dx∫Put x2 = t 2x dx = dt

I =t1

2t e dt∫

=t t1

–2

dtt e dt e dt

dt

∫ ∫ ∫

=t t1–

2t e e dt ∫

=t t1– +

2t e e c

= [ ]t1–1 +

2e t c

=

22

x

x –1 +2

ee c



Q-15) . sin∫∫∫∫ xe x dx

Ans. Let I =x .sine x dx∫

=x xsin – . sin

dx e dx e dx x dx

dx

∫ ∫

=x xsin . – cosx e e x dx∫

=

( )

x

x x

sin .

– cos – – sin

x e

x e dx e x dx ∫ ∫

∴∴∴∴ I =x x x.sin – cos. – sine x e e dx∫

∴∴∴∴ I = ex(sin x – cos x) – I

∴∴∴∴ 2I = ex(sin x – cos x)

∴∴∴∴ I =

x

2

e(sin x – cos x) + c

Q-16) cos (2 log ) x dx∫∫∫∫Ans. Put 2logx = t

∴∴∴∴ log x2 = t

∴∴∴∴ 2

t

e = x

2

x dx = dt

dx =2

dt · 2

t

e

I = 2

t1

cos ·2

t e dt∫

I =2 2

t t1

cos – (cos ) 2

dt e dt e dt t dt

dt

∫ ∫∫

= ( )2 21

t t1

2cos · – 2 · – sin +2

t e e t dt c ∫

= ( )2 2 2

t t t

cos + sin – sind

t e t e dt e dt tdt

∫ ∫

= cos t · 2

t

e + 2 sint 2

t

e

2

t

–2 cos +e t dt c∫ I = cos t · 2

t

e + 2 sint 2

t

e – 4I + c

I =5

x [cos(2 log x) + 2 sin (2 log x)] + c

Q-14)3sec∫∫∫∫ x dx

Ans. Let I =3sec x dx∫

I = ( )2sec secx x dx∫

=

( )

2

2

sec sec

– sec sec

x x dx

dx dx dx

dx

∫∫ ∫

= sec x tan x

– 1sec tan .tan +x x x dx c∫= sec x tan x

( )2

1– sec sec –1 +x x dx c∫= sec x tan x

( )2

1– sec sec –1 +x x dx c∫= sec x tan x

3

1– sec + sec +x dx x dx c∫ ∫2I = sec x tan x + log|sec x + tan x| + c

2

I = 1

2[sec x tan x + log|sec x + tan x|+ c

GROUP (K) – HOME WORK PROBLEMS

Q-1)2secx x dx∫∫∫∫

Ans. =2 2sec – sec ( )

dx x dx x dx x dx

dx

∫ ∫ ∫

= 1tan – tan · 1 +x x x dx c∫= x tanx – log |secx| + c

= x tanx + log|cosx| + c



Q-2) sinx x dx∫∫∫∫Ans. Let = sinx x dx∫

= sin – sin ·d

x x dx x dx x dxdx

∫ ∫ ∫

= x (– cos x) – – cos .1x dx∫= –x cos x + sin x + c

Q-3)2 logx x dx∫∫∫∫

Ans. Let =2 logx x dx∫

Here neither an exponential nor a

trigonometric function is present we take

the algebric function x2 as

∴∴∴∴ v dx∫ =2 log .x x dx∫

∴∴∴∴ ( )2log .x x dx∫=

3 3 1log – . +

3 3

x xx dx c

x

∫

=3 2

1

1 1log – +

3 3x x x dx c∫

=3 3

1

1 1log – +

3 9x x x c

Q-4)2.cosx x dx∫∫∫∫

Ans. Let I =2.cosx x dx∫

= cos – cosd

x x dx x dx x dxdx

∫ ∫ ∫2 2

= ( ).sin – sin 2x x x x dx∫2

= 2.sin – 2

sin – sin

x x

dx x dx x dx x dx

dx

∫ ∫ ∫

= x2. sin x – 2 – cos + cos .1x x x dx ∫

= x2. sin x + 2 x cos x – 2 sin x + c

Q-5)1 – cos2

xdx


1– cos2

xdx

x∫= 22sin

xdx

x∫=

21.cos

2x ec x dx∫

=

( )

2

2

cos1

2– cos .

x ec x dx

dec x dx x dx

dx

∫∫

= ( ) ( )1

– cot – – cot 12x x x dx ∫

=1

2[–x . cot x + log |sin x|] + c.

Q-6)2 –1cosx x dx∫∫∫∫

Ans. Let I =2 –1cosx x dx∫

= ( )–1 2cos .x x dx∫= ( )

( )

–1

2 –1 2

cos

– cos .

x

dx dx x x dx dx

dx

∫ ∫ ∫

= ( )3 3

–1

2

–1cos – .

3 31+

x xx dx

x∫

=

3 –1 3

2

cos 1+

3 3 1+

x x xdx

x∫

=

3 –1 2

2

cos 1+ .

3 3 1 –

x x xx dx

x∫Put 1 – x2 = t

∴∴∴∴ –2x dx = dt

∴∴∴∴ x dx =–

2

dtand x2 = 1 – t

∴∴∴∴ I =( )3 –1 –cos 1 1 –

+ .3 3 2

dtx x t

t∫=

3 –1cos 1 1 ––

3 6

x x tdt

t∫



Q-7)–1sin xdx∫∫∫∫

Ans. =–1sin .1x dx∫= –

duuvdx u vdx vdx dx

dx

∫ ∫ ∫ ∫

∴∴∴∴ I = sin–1 x.x – 2

+ ,1 –

xdx c

x∫∴∴∴∴ 1 – x2 = t ∴∴∴∴ – 2x dx = dt

∴∴∴∴ x dx = –

2

dt

I = x sin–1 x – 1 –1

+2dt c

t

∫

= x sin–1 x + ( )12 +

2t c

= x sin–1x+ 21 – +x c

Q-8) sin xdx∫∫∫∫Ans. =x t

∴∴∴∴ x = t 2

∴∴∴∴ dx = 2t dt

I = sin .2t tdt∫ = 2 sint tdt∫= ( ) ( )2 cos – cos .1 +t t t dt c

∫

= 2 [– t cost + sin t] + c

= –2 cos + sin +x x x c

Q-9)–1tan x dx∫∫∫∫

Ans. Let I =–1tan x dx∫

Put = t ∴∴∴∴ x = t2 ∴∴∴∴ dx = 2t dt

∴∴∴∴ I = ( ) ( )–1tan 2t t dt∫= ( )

( )

–1

–1

tan . 2

– tan . 2

t t dt

dt t dt dt

dt

∫∫ ∫

= t2 tan–1 t –2

2

1.

1+t dt

t∫= t2 tan–1 t –

( )2

2

1+ –1

1+

tdt

t∫= t2 tan–1 t – 2

1 +1+

dtdt

t∫ ∫= t2 . tan–1 t – t + tan–1 t + c

= (t2 + 1)tan–1 t – t + c

= (x + 1)tan–1 – + .x x c

Q-10) (((( ))))cos . log sinx x dx∫∫∫∫Ans. Let I = ( )cos .log sinx x dx∫

Put sin x = t


∴∴∴∴ I = log t dt∫= ( )log .1t dt∫= log 1 – 1 log

dt dt dt t dt

dt

∫ ∫ ∫= t. log t

1– t dt

t

∫

= t. log t – dt∫ = t. log t – t + c

= t (log t – 1) + c = sin x [log (sin x)–1] + c

Q-11) .cosxe x dx∫∫∫∫

Ans. Let I =x .cose x dx∫

=

3 –1cos 1 1– –

3 6

x xt dt

t

∫

=

1 13 –1–2 2

cos 1 1– +

3 6 6

x xt dt t dt∫ ∫

=

1 3–

3 –1 2 2cos 1 1– . + +

1 33 6 6

2 2

x x t tc

= ( ) ( )1 33

–1 2 22 21 1

cos – 1 – + 1 – +3 3 9

xx x x c

=( )

( )

32 2 13

–1 2 21 –1

cos + – 1 – +3 3 3

xxx x c



Q-12) (((( ))))sin log x dx∫∫∫∫Ans. Let I = ( )sin log x dx∫

Put log x = t

∴∴∴∴ x = et

∴∴∴∴ dx = et dt

∴∴∴∴ I =tsin .t e dt∫

=t tsin . – sin

dt e dt e dt t dt

dt

∫ ∫ ∫= et. sin t t– cose t dt∫= ( )t t t.sin – cos – – sine t t e dt e t dt

∫ ∫

∴∴∴∴ I = et. sin t – cos t. et – I + c1

∴∴∴∴ 2I = et (sin t – cos t) + c1

∴∴∴∴ I =

t

2

e(sin t – cos t) +

1

2

c

∴∴∴∴ I =

t

2

e(sin t – cos t) + c, Where c =

1

2

c

Q-13)3cosec x dx∫∫∫∫

Ans. Let I =3cosec x dx∫

=2cos .cosec x ec x dx∫

Q-14) ∫∫∫∫ 3sec 4x dx

Ans. Let I = sec 4∫ 3x dx

=

( )

sec 4 sec 4

– sec 4 sec4

∫

∫ ∫

2

2

x x dx

dx dx x dx

dx

=

( )

tan4sec 4 .

4

tan4– .sec 4 .tan4 4 .

4

∫

xx

xx x dx

= ( )sec4 .tan4

– sec 4 –1 .sec4 .4 ∫ 2x x

x x dx

=sec 4 .tan4

– sec 4 + sec 44 ∫ ∫3x x

x dx x dx

I =sec4 .tan4

–4

1+ log sec4 + tan4 +

4×××× 1

x xI

x x c

2I =log sec4 + tan4sec 4 .tan4

+ +4 4

1

x xx xc

∴∴∴∴ I = 1

8[sec 4x.tan 4x + log|sec 4x + tan 4x|] + c

=

( )

2

2

cos cos

– cos . cos

ec x ec x dx

dec x ec x dx dx

dx

∫ ∫∫ ∫

= (cosec x)(– cot x)

( ) ( )– – cos .cot – cotec x x x dx∫= – cosec x cot x

2– cos .cotec x xdx∫= – cosec x cot x ( )2– cos cos –1ec x ec x dx∫= – cosec x cot x

3– cos + cosec dx ec x dx∫ ∫∴∴∴∴ I = – cosec x cot x – I + log|cosec x – cot x|

∴∴∴∴ 2I = – cosec x cot x + log|cosec x – cot x|

∴∴∴∴ I =– cos cot 1

+2 2

ec x x

log|cosec x – cot x| + c

or I =– cos .cot 1

+ log tan +2 2 2

ec x x xc

=x xcos – . cos

dx e dx e dx x dx

dx

∫ ∫

= ( )x xcos . – – sinx e e x dx∫= x

x x

cos .

+ sin – cos

x e

x e dx e x dx ∫ ∫

∴∴∴∴ I = cos.ex + sin x.ex – I + c1

∴∴∴∴ 2I = ex(sin x + cos x) + c1

∴∴∴∴ I =x

2

e(sin x – cos x) +

1

2

c

∴∴∴∴ I =x

2

e(sin x – cos x) + c, where c =

1

2

c



GROUP (L) – CLASS WORK PROBLEMS

Q-1) (((( ))))2cot – cot – 1xe x x dx∫∫∫∫

Ans. Let I = ( )2x cot – cot –1e x x dx∫= ( )2x cot – cose x ec x dx∫

Put f (x) = cot x. Then

f ′′′′(x) = – cosec2 x

∴∴∴∴ I = ( ) ( )'x +e f x f x dx ∫= ex f(x) + c

= ex cot x + c

Q-2)(((( ))))

2

1+

2 +

x xe dx

x∫∫∫∫

Ans. Let I = ( )2

x 1+

2+

xe dx

x∫

I =( )

( )2

x2+ –1

2+

xe dx

x∫

I =( )

2

x 1 1–

2+ 2+e dx

x x

∫

I =( )

( )2

x–11

+2+ 2+

e dxx x

∫

Put f(x) =1

2+ x

Then f ′′′′(x) = ( )2

–1

2+ x

So,

I = ( ) ( )'x +e f x f x dx ∫∴∴∴∴ I = ex f(x) + c

∴∴∴∴ I =x

+2+

ec

x

Q-3)1 – sin

1 – cos

x xe dx

x

∫∫∫∫

Ans. Let I =x 1– sin

1 – cos

xe dx

x

∫

I = 2

x

1 – 2sin .cos2 2

2sin2

x x

e dxx

∫

I =2x1

cos – 2cot2 2 2

x xe ec dx

∫

I =2x1

–2cot + cos2 2 2

x xe ec dx

∫

put f(x) = –2 cot 2

x

than f ′′′′(x) = cosec2

2

x

so,

I = ( ) ( )'x1+

2e f x f x ∫

I = ( )x1+

2e f x c

I =x1–2cot +

2 2

xe c

⇒⇒⇒⇒ I =x– cot +

2

xe c

Q-4)

(((( ))))

2

32 2

– +1

+1

x x xe dx

x∫∫∫∫

Ans. Let I =

( )

2

32 2

x – +1

+1

x xe dx

x∫

I =( )

( )

2

32 2

x+1 –

+1

x xe dx

x

∫

I =( )

( ) ( )

2

3 32 22 2

x+1

–

+1 +1

x xe dx

x x

∫



I =

( ) ( )1 3

2 22 2

x 1–

+1 +1

xe dx

x x

∫

f(x) =

( )( )

1–

2 21

2 2

1= +1

+1

x

x

Then f ′′′′(x) = ( ) ( )3

–2 22

1– +1 . +12

dx x

dx

=

( )3

2 2

2–

2 +1

x

x

=

( )3

2 2

–

+1

x

x

So,

I = ( ) ( )'x +e f x f x dx ∫I = ex f(x) + c

=

( )1

2 2

x 1+

+1

e c

x

I = 2

x

++1

ec

x

Q-5) (((( )))) (((( ))))sin log + cos logx x dx ∫∫∫∫

Ans. Let I = ( ) ( )sin log + cos logx x dx ∫Put x = t ⇒⇒⇒⇒ x = et ⇒⇒⇒⇒ dx = et dt

I = [ ] tsin + cost t e dt∫Put f (t) = sin t ⇒⇒⇒⇒ f ′′′′(t) = cos t

I = ( ) ( )'t +e f t f t dt ∫= et f (t) + c

= et sin t + c

I = x sin (log x) + c

Q-6)(((( ))))2

–1tan

2

1+ +

1+

xx x

e dxx∫∫∫∫

Ans. Let I = ( )2

-1tan

2

x1+ +

1+

x xe dx

x∫Put tan–1 x = t ⇒⇒⇒⇒ x = tan t

∴∴∴∴ 2

1=

1+

dxdt

x

I = ( )2t 1+ tan + tane t t dt∫I = ( )2t sec + tane t t dt∫I = ( )2t tan +sece t t dt∫Let f(t) = tan t ⇒⇒⇒⇒ f ′′′′(t) = sec2 t

I = ( ) ( ){ }'t +e f t f t dt∫So,

I = et f (t) + c

I =-1tan x +e x c [ ]tan =t x∵

I =-1tan x +xe x c

Q-7) (((( ))))2

log

1+ log

xdx

x∫∫∫∫

Ans. Let I = ( )2

log

1+ log

xdx

x∫Put log x = t

Then x = et ⇒⇒⇒⇒ dx = etdt

I = ( )2

t

1+

te dt

t∫

I =( )

( )2

t1+ –1

1+

te dt

t

∫

I =( )

2

t 1 1–

1+ 1+e dt

t t

∫

If ( )1

=1+

f tt



Then ( )( )

2'

–1=1+

f tt

I = ( ) ( )'t +e f t f t dt ∫I = et f(t) + c

I =t 1. +1+

e ct

I = +1+ log

xc

x

Q-8)(((( ))))

2

2

1

1

x xe dx

x

++++

++++ ∫∫∫∫

Ans. I =( )

2

2

x 1+

1+

xe dx

x

∫

=( )

2

2

x +1+ 2 – 2

+1

x x xe dx

x

∫

=( )

( )

2

2

+1 – 2

+1

xx x

e dxx

∫

=( )

2

x x– 2+1

xe dx e dx

x∫ ∫

=( )

( )x x

+1 –1– 2

+1

xe e dx

x∫

I =( )

2

x x1 –1– 2 +

+1 +1e e dx

x x

∫

Let ( ) ( )( )

1 –1= =

+1 +1f x f x

x x′

2

I = ( ) ( )( )– 2 +e e f x f x dx

′ ∫x x

= ( )x x– 2 +e e f x c

=

xx 2– +

+1

ee c

x

=x–1+

+1

xe c

x

I =x –1

++1

xe c

x

Q-9)(((( ))))

2

2

log –1

1 log

xdx

x

++++ ∫∫∫∫

Ans. I =( )

2

2

log –1

1+ log

xdx

x

∫

Put log x = t ∴∴∴∴ x = et


∴∴∴∴ I =

2

2

t–1

1+

te dt

t

∫

=( )

2

22

t – 2 +1

1+

t te dt

t

∫

=

( )

( ) ( )

2

2 22 2

t1+ 2

–1+ 1+

t te dt

t t

∫

=( )

2 22

t 1 2–

1+ 1+

te dt

t t

∫

Now, 2

1

1+

d

dt t

= ( )–1

21+d

tdt

= ( ) ( ) ( )–1–1

2 2–1 1+ 1+d

t tdx

= ( ) ( ) ( )–2

2–1 1+ 0 + 2t t

=( )

22

–2

1+

t

t

∴∴∴∴ I = 2 2

t 1 1+

1+ 1+

de dt

dtt t

∫

= 2

t 1+

1+e c

t

=( )

2

1+

1+ logx c

x

I = ( )2+

1+ log

xc

x



Q-6)

2

2 2

( – 1)

( +1)

x xe dx

x∫∫∫∫

Ans. =

2

4 2

x – 2 +1

+ 2 +1

x xe dx

x x

∫

=

2

2 2 2 2

x +1 2–

( +1) ( +1)

x xe dx

x x

∫

= 2 2 2

x 1 2–

1 ( +1)

xe dx

x x

∫ +

Let, f(x) = 2

1

+1x

f′ (x) =( )

2

–1 2

+1

x

x

( ) ( ) ( )'x x∴∴∴∴ + = +e f x f x dx e f x c

∫= 2

x

+1

e

x + c

GROUP (L) – HOME WORK PROBLEMS

Q-1) (sin + cos ) xe x x dx∫∫∫∫Ans. Let, f(x) = sin x

∴∴∴∴ f′ (x) = cos x

= exsinx + c

( ) ( ) ( )'x x∴∴∴∴ + = +e f x f x dx e f x c

∫

Q-2) 2

cos + sin

cos

x x xe dx

x∫∫∫∫Ans. I = ( )x sec + tan sece x x x∫

Let, f(x) = sec x

∴∴∴∴ f′ (x) = tan x.,

I = = ex sec x + c

( ) ( ) ( )'x x∴∴∴∴ + = +e f x f x dx e f x c

∫

Q-3) 2 + sin2

1+ cos2

x xe dx

x

∫∫∫∫

Ans. I = 2

x 2+2sin cos

2cos

x xe dx

x

∫

= ( )2x sec + tane x x dx∫Let, f(x) = tan x

f′ (x) = sec2 x

I = ex tanx + c

( ) ( ) ( )+ = +e f x f x dx e f x c

∫ 'x x∵∵∵∵

Q-4) 2 – sin2

1 – cos2

x xe dx

x

∫∫∫∫

Ans. I = 2

x 2 – 2sin cos2

2sin

x xe dx

x

∫

= ( )2x cos – cote ec x x dx∫Let, f(x) = – cot x

f′ (x) = cosec2 x

I = – ex cot x + c

( ) ( ) ( )+ = +e f x f x dx e f x c

∫ 'x x∵∵∵∵

Q-5) 1+ sin2

41+ cos2

x xe dx

x

∫∫∫∫

Ans. put 2x = t

∴∴∴∴ 2 dx = dt

Let, I = 2x 1+ sin2

1+cos2

xe dx

x

∫

=t ××××1+sin

1+cos 2

t dte

t

∫

=2 2

t

2sin cot1 1 2 2 +2

2cos 2cos2 2

t t

e dtt t

∫

=2t1 1

sec + tan2 2 2 2

t te dt

∫

Let, f(x) = tan 2

t

f′ (x) = sec2 1.

2 2

t

( ) ( ) ( )'x x∴∴∴∴ + = . +e f x f x dx e f x c

∫=

t1tan +

2 2

te c

=2x1tan +

2e x c



Q-7) 2( +1)

xxedx

x∫∫∫∫

Ans. = 2 2

x +1 1–

( +1) ( +1)

xe dx

x x

∫

= 2

x 1 1–

+1 ( +1)e dx

x x

∫

Let, f(x) = 1

1+ x

f′ (x) = ( )2

–1

1+ x

∴∴∴∴ I =x

+1

e

x + c

Q-8) 1

+ logxe x dxx

∫∫∫∫

Ans. put f(x) = log x

∴∴∴∴ f’ (x) = 1

x

∴∴∴∴ I = ( ) ( )'x +e f x f x dx ∫= ex f(x) + c

I = ex log x + c

Q-9) (tan – log cos )xe x x dx∫∫∫∫Ans. f(x) = –log(cos x)

∴∴∴∴ f′(x) =–1

cos x ·(–sinx) = tanx

( ) ( ) ( )'x x+ = +e f x f x dx e f x c ∫ = – ex (log cosx) + C

Q-10) (((( ))))cot + log sinxe x x dx∫∫∫∫

Ans. Let I = ( )x cot + log sine x x dx∫Put f(x) = log sinx. Then

f ′(x) = d

dx(log sin x) =

1

sinx. cos x = cot x

∴∴∴∴ I = [ ]'x ( )+ ( )e f x f x dx∫= ex.f(x) + c

= ex.log sin x + c

Q-11) (((( ))))sin sin cos cosxe x x x dx++++∫∫∫∫

Ans. I = ( )sin x cos sin +1e x dx∫put sinx = t ∴∴∴∴ cos x dx = dt

∴∴∴∴ I = ( )t +1e t dt∫f ′ (t) = t f ′ (t) = 1

I = et . t + c

I = esinx . sin x + c

Q-12)2

–1sin

2

1 –

1 –

x x xe dx

x

++++ ∫∫∫∫

Ans. Put sin–1 x = y ∴∴∴∴ x = sin y

∴∴∴∴ 2

1

1 – xdx = dy

I =2y sin + 1 – sine y y dy

∫= [ ]y sin + cose y y dy∫

f (y)= sin y f ′ (y) = cos y

I = [ ]y sin + cose y y dy∫= ey sin y + c

I = esin–1x sin (sin–1 x) + c

I = esin–1x (x) + c

GROUP (M) – CLASS WORK PROBLEMS

Q-1) 2

1

– 3 + 2dx

x x∫∫∫∫

Ans. Let I = 2

1

– 3 +2dx

x x∫

= ( ) ( )

1

– 2 –1∫ dxx x ...(i)

Let ( ) ( )

1

– 2 –1∫ dxx x

= +–1 – 2

A B

x x



Q-4) 2

sec2

cos + 4cos + 3

xdx

x x∫∫∫∫Ans. Let I = 2

sec2

cos + 4cos +3

xdx

x x∫= 2

2sin .cos

cos + 4cos +3

x xdx

x x∫Put cos x = t

∴∴∴∴ sinx dx = –dt

∴∴∴∴ I = 2

–2

+ 4 +3

t dt

t t∫= ( ) ( )

–2+ 3 +1

t dt

t t∫

Let ( ) ( )+ 3 +1

t

t t

= ++ 3 +1

A B

t t

∴∴∴∴ A == –3t+1

t

t

=–3

–3 +1=

3

2

B == –1t+ 3

t

t

=–1

–1+ 3=

1–2

∴∴∴∴ I =

–13

22–2 ++3 +1

dtt t

∫

=3 1

–2 log + 3 – log +1 +2 2

t t c

= log|cos x + 1| – 3 log|cos x + 3| + c

Q-2) 2

3 – 4

– 3 + 2

xdx

x x∫∫∫∫

Ans. Let I = 2

3 – 4

– 3 +2

xdx

x x∫

I = ( ) ( )

3 – 4

–1 – 2

xdx

x x∫

Let ( ) ( )

3 – 4

–1 – 2

x

x x = +–1 – 2

A B

x x

∴∴∴∴ 3x – 4 = A (x – 2) + B(x – 1)

Put x = 1 ⇒⇒⇒⇒ – 1 = A (–1) ⇒⇒⇒⇒ A = 1

Put x = 2 ⇒⇒⇒⇒ – 2 = B (1) ⇒⇒⇒⇒ B = 2

I = ( ) ( )

3 – 4

–1 – 2

xdx

x x∫

I = 1 2

+–1 – 2

dxx x

∫

I = log|x – 1| + 2 log |x – 2| + c

Q-3) (((( )))) (((( ))))

2sec

2 + tan 3 + tan

xdx

x x∫∫∫∫

Ans. Let I = ( ) ( )

2sec

2+ tan 3+ tan

xdx

x x∫Put tan x = t


∴∴∴∴ I = ( ) ( )2+ 3 +

dt

t t∫ ... (i)

Let ( ) ( )2+ 3+

dt

t t

= +2+ 3 +

A B

t t... (ii)

∴∴∴∴ A ==1x

1= –1

– 2x

B ==2x

1=1

–1x

∴∴∴∴ I =–1 1

+–1 – 2

∫ dxx x

= –log|x – 1| + log |x – 2| + c

∴∴∴∴ I =– 2

log +–1

xc

x

∴∴∴∴ A == –2t

1=1

3+ t

B == –3t

1= –1

2+ t

∴∴∴∴ A = 1 and B = – 1 ...(iii)

∴∴∴∴ From (i), (ii), and (iii)

∴∴∴∴ I =( )–11

+2+ 3+

dtt t

∫

= log|2 + t| – log |3 + t| + c

=2+

log +3+

tc

t

=2+ tan

log +3+ tan

xc

x



Q-5)

2

2

–

– 2 – 3

x xdx


2

2

–

– 2 – 3

x xdx

x x∫= ( )

2

2

––1

– 2 – 3

x xdx

x x∫= ( )

2

2

– 2 + – 3 +3–1

– 2 – 3

x x xdx

x x∫= ( )

2

2 2

– 2 – 3 +3–1 –

– 2 – 3 – 2 +3

x x xdx dx

x x x x∫ ∫= ( )

2

1 2 +6–1 –

2 – 2 +3

xdx dx

x x∫ ∫= ( )

2

1 2 – 2+8–1 –

2 – 2 – 3

xdx dx

x x∫ ∫= ( )

2

2

1 2 – 2–1 –

2 – 2 – 3

1– 8

– 2 – 3

xx dx

x x

dxx x

∫∫

= 2

2

1– – log – 2 – 3

2

1– 8

– 2 +1 –1 – 3

x x x

dxx x∫

=( ) ( )

2

2 2

1– – log – 2 – 3 – 8

2 –1 – 2

dxx x x

x∫

=2

××××××××

1– – log – 2 – 3

2

1 –1 – 2– 8 log +

2 2 –1+ 2

x x x

xc

x

∴∴∴∴ I =21 – 3

– – log – 2 – 3 – 2log +2 –1

xx x x c

x

Q-6)

2

2 2

+10 + 21

xdx


2

4 2

+10 + 21

xdx

x x∫ ... (i)

Put x2 = t

∴∴∴∴

2

4 2+10 +21

x

x x= 2 +10 + 21

t

t t

= ( ) ( )+ 7 + 3

t

t t

= ++ 7 + 3

A B

t t

∴∴∴∴ A == –7t+3

t

t

=–7

–4

=7

4

∴∴∴∴ B == –3t+ 7

t

t

=–3

–3 + 7

=–3

4

∴∴∴∴

2

4 2+10 +21

x

x x=

7 –3

4 4++ 7 +3t t

=2 2

7 –3

4 4++ 7 +3x x

... (ii)

∴∴∴∴ From (i) and (ii),

∴∴∴∴ I = 2 2

7 3

4 4–+ 7 +3

dxx x

∫

= 2 2

7 1 3–

4 4+7 +3

dxdx

x x∫ ∫=

( ) ( )2 2

2 2

7 3–

4 4+ 7 + 3

dx dx

x x∫ ∫

= –1

–1

××××

××××

7 1tan

4 7 7

3 1– tan +4 3 3

x

xc

∴∴∴∴ I =–1 –17 3

tan – tan +4 47 3

x xc

Q-7) (((( )))) (((( ))))

2

2 2

2 –1

+ 4 – 5

xdx


( ) ( )

2

2 2

2 –1

+ 4 – 5

xdx

x x∫ ... (i)

Put x2 = t

∴∴∴∴( ) ( )

2

2 2

2 –1

+ 4 – 5

x

x x= ( ) ( )

2 –1

+ 4 – 5

t

t t

= ++ 4 – 5

A B

t t

∴∴∴∴ A == –4t

2 –1=1

– 5

t

t



Q-8)(((( )))) (((( ))))

2– 1 +1

dx


( ) ( )2

–1 +1

dx

x x∫ ... (i)

Let

( ) ( )2

1

–1 +1x x= ( ) ( ) ( )2

+ +–1 +1–1

A B C

x xx

=( ) ( ) ( ) ( )

( ) ( )

2

2

–1 +1 + +1 + –1

–1 +1

A x x B x C x

x x

⇒⇒⇒⇒ A(x2 – 1) + B (x + 1) + C(x – 1)2 = 1

Put x = 1

∴∴∴∴ 2B = 1

∴∴∴∴ B = 1

2

Put x = –1

∴∴∴∴ 4C = 1

∴∴∴∴ C = 1

4

Put x = 0

∴∴∴∴ –A + 1

2 + 1

4 = 1

∴∴∴∴ A = 1–4

∴∴∴∴( ) ( )

2

1

–1 +1x x=

( ) ( ) ( )2

11 1–

24 4+ +–1 +1–1x xx

.... (ii)

Q-9)(((( )))) (((( ))))

2

2

5

+1 + 4

xdx

x x∫∫∫∫

Ans. Let I = ( ) ( )

2

2

5

+1 + 4

xdx

x x∫ ... (i)

Let

( ) ( )

2

2

5

+1 + 4

x

x x =( ) ( )2

++

–1 + 4

A Bx c

x x

... (ii)

( ) ( )

2

2

5

+1 + 4

x

x x = ( ) ( ) ( )

( ) ( )

4 + + +1

+1 + 4

A x Bx C x

x x

+2

2

∴∴∴∴ A(x2 – 4) + (Bx + C)(x – 1) = 5x2

Put x = –1,

∴∴∴∴ 5A + (C – B)(O) = 5

∴∴∴∴ A = 1

Put x = 0,

∴∴∴∴ 4A + C = 0

∴∴∴∴ C = –4A

= –4(1) = –4

Put x = 1

∴∴∴∴ 5A + 2(B + C) = 5

∴∴∴∴ 5(1) + 2(B – 4) = 5

∴∴∴∴ 2(B – 4) = 0

∴∴∴∴ B = 4

∴∴∴∴( ) ( )

2

2

5

+1 + 4

x

x x= 2

1 4 + 4+

–1 + 4

x

x x

.... (iii)

∴∴∴∴ From (i) and (iii)

∴∴∴∴ I = 2

1 4 – 4+

+1 – 4

xdx

x x

∫

=

( ) ( )

1 2+2

+1 – 4

1– 4

+ 2

xdx dx

x x

dxx

∫ ∫∫

2

22

∴∴∴∴ B ==5t

2 –1=1

+ 4

t

t

∴∴∴∴( ) ( )

2

2 2

2 –1

+ 4 – 5

x

x x=

1 1+

+ 4 – 5t t

= 2 2

1 1+

+ 4 – 5x x... (ii)

∴∴∴∴ From (i) and (ii),

∴∴∴∴ I = 2 2

1 1+

+ 4 +5dx

x x

∫

= ( ) ( )2 22 2

++ 2 – 5

dx dxdx

x x∫ ∫

=–11 1 – 5

tan + log +2 2 2 5 + 5

x xc

x


∴∴∴∴ I =( ) ( ) ( )2

1–1 1

24 4+ +–1 –1–1

dxx xx

∫

= ( )

1 1 1– log –1 – + log +1 +4 2 –1 4

x x cx

= ( )

1 +1 1log – +

4 –1 2 –1

xc

x x



Q-10) (((( ))))

1

cos 2 + sindx

x x∫∫∫∫Ans. Let I = ( )cos 2 + sin

dx

x x∫ ... (i)

Put sin x = t


∴∴∴∴ dx = cos

dx

x.

∴∴∴∴ I =( )

cos

cos 2+

dt

x

x t

∫

= ( )2cos 2+

dt

x t∫

= ( ) ( )21 – sin 2+

dt

x t∫

= ( ) ( )21 – 2+

dt

t t∫

= ( ) ( ) ( )1+ 1 – 2+

dt

t t t∫ ... (ii)

Let ( ) ( ) ( )

1

1+ 1 – 2+t t t

= + +1+ 1 – 2+

A B C

t t t

∴∴∴∴ A = ( ) ( )= –1t

1

1 – 2+t t

=1

2

B = ( ) ( )=1t

1

1+ 2+t t

=1

6

C = ( ) ( )= –2t

1

1+ 1 –t t

=1–3

∴∴∴∴ ( ) ( ) ( )

1

1+ 1 – 2+t t t=

1 1 –1

2 6 3+ +1+ 1 – 2+t t t

... (iii)

∴∴∴∴ I = log|x – 1| + 2 log|x2 + 4|

– 4 × × × × –11

tan +2 2

xc

= log|x – 1| + 2 log|x2 + 4|

– –12tan + .2

xc

∴∴∴∴ From (ii) and (iii)

∴∴∴∴ I =

1 1 –1

2 6 3+ +1+ 1 – 2+

dtt t t

∫

=1 1 1log 1+ – log 1+ – log 2+ +

2 6 3t t t c

∴∴∴∴ I =1 1log 1+ sin – log 1 – sin

2 6

1– log 2+ sin +3

x x

x c

GROUP (M) – HOME WORK PROBLEMS

Q-1) 2

1

– 5 + 6dx


1

– 5 +6dx

x x∫= ( ) ( )

1

– 3 – 2dx

x x∫ ...(i)

Let ( ) ( )

1

– 3 – 2x x∫

= ( ) ( )+

– 3 – 2

A B

x x

∴∴∴∴ A ==3x

1=1

– 2x

∴∴∴∴ B ==2x

1= –1

– 3x

∴∴∴∴( ) ( )

1

– 3 – 2x x∫ =( ) ( )

1 1–

– 3 – 2x x

....(ii)


∴∴∴∴ I =( )

( )

–11+

– 3 – 2dx

x x

∫

=1 1

–– 3 – 2

dx dxx x∫ ∫

= –log|x – 3| + log |x – 2| + c

=– 3

log +– 2

xc

x



Q-2) 2

3 – 2

– 3 + 2

xdx

x x∫∫∫∫Ans. Let = ( ) ( )

3 – 2

– 2 –1

xdx

x x∫3x – 2 = A(x – 1) + B(x – 2)

x = 2, x = 1

A = 4; –B = 1

B = – 1

=4 –1

+– 2 –1dx dx

x x∫ ∫= 4 log |x – 4| – log |x – 1| + c

Q-3) (((( )))) (((( ))))

log

2 + log 3 + logdx

x x x∫∫∫∫Ans. put log x = t

1

xdx = dt

+2+ 3 +

A B

t t = ( ) ( )2+ 3+

t

t t

A (3 + t) + B(2 + t) = t

put t = –2, t = –3

A = – 2 B = 3

I =–2 3

+2+ 3+

dt dtt t∫ ∫

I = –2log 1+ +3log 3+t t

I = 3 log |3 + log x| – 2log |2 + log x| + c

=( )

( )

3

2

3 + loglog +

2+ log

xc

x

Q-4) 2

sin2

sin – 5sin + 6

xdx


sin2

sin – 5sin +6

xdx

x x∫= 2

2sin .cos

sin – 5sin +6

x x dx

x x∫put sinx = t


∴∴∴∴ I = 2

2

– 5 +6

t dt

t t∫=

( ) ( )

2

– 3 – 2

tdt

t t∫ ...(i)

Let ( ) ( )

2

– 3 – 2

tdt

t t∫

Q-5)

2

2

+1

– 4 + 3

xdx

x x∫∫∫∫Ans. Let =

2

2

+1

– 4 + 3

xdx

x x∫

2 2

2

+ +

1

– 4 – 3 +1

– 4 –3

4 – 2

x x x

x x

x

∴∴∴∴ I = 2

4 – 21+

– 4 +3

xdx

x x

∫

∴∴∴∴ I = 2

4 – 2+

– 4 +3

xdx dx

x x∫ ∫ ....(i)

4x – 2 = Ad

dx(x2 – 4x + 3) + B

= A(2x – 4) + B .....(ii)

= 2Ax – 4A + B

∴∴∴∴ 2A = 4

∴∴∴∴ A = 2

∴∴∴∴ –4A + B = – 2 ⇒⇒⇒⇒ B = 6

∴∴∴∴ 4x – 2 = 2(2x – 4) + 6 .....(iii)


∴∴∴∴ I =( )2

2 2 – 4 +6+

– 4 + 3

xdx dx

x x

∫ ∫= 2 2

2 – 4+2 +6

– 4 +3 – 4 +3

x dxx dx

x x x x∫ ∫=

2

2+2log – 4 +3 +6

– 4 + 4 –1

dxx x x

x x∫= x + 2 log |x2 – 4x + 3|

××××1 – 2 –1

+6 log +2 – 2+1

xc

x

= ( ) ( )+

– 3 – 2

A B

t t

∴∴∴∴ A ==3t

2= 6

– 2

t

t

∴∴∴∴ B ==2t

2= – 4

– 3

t

t

∴∴∴∴( ) ( )

2

– 3 – 2

t

t t=

( )

( )

( )

–46+

– 3 – 2t t

∴∴∴∴ I =6 4

–– 3 – 2

dtt t

∫

= 6log|t – 3| – 4 log |t – 2| + c

= 6log|t – 3| – 4 log |sinx – 2| + c



∴∴∴∴ I = x + 2 log |x2 – 4x + 3|

+ 3 log – 3

–1

x

x+ c

∴∴∴∴ I = x + 2 log |(x – 3)(x –1)| + 3log – 3

–1

x

x+ c

∴∴∴∴ I = x + log ( ) ( )( )

( )

3

2 2

3××××

– 3– 3 –1 +

–1

xx x c

x

∴∴∴∴ I = x + log ( )

( )

5– 3

+–1

xc

x

∴∴∴∴ I = x + 5 log |x – 3| – log|x – 1| + c

Q-6) (((( )))) (((( ))))

2

2 2

+ 4 + 9

xdx


( ) ( )

2

2 2

+ 4 + 9

xdx

x x∫ ... (i)

Let x2 = t

∴∴∴∴( ) ( )

2

2 2+ 4 + 9

x

x x=

( ) ( )+ 4 + 9

t

t t

= ++ 4 +9

A B

t t

∴∴∴∴ A == –4t+ 9

t

t

=4–5

∴∴∴∴ B == –9t+ 4

t

t

=9

5

∴∴∴∴( ) ( )

2

2 2+ 4 + 9

x

x x=

( ) ( )+ 4 + 9

t

t t

= ++ 4 + 9

A B

t t

∴∴∴∴ A ==4t

4= –

+ 9 5

t

t

∴∴∴∴ B == –9t

9=

+ 4 5

t

t

∴∴∴∴ ( ) ( )

2

2 2+ 4 + 9

x

x x =

–4 9

5 5++ 4 + 9t t

=2 2

–4 9

5 5++ 4 + 9x x

...(iii)


∴∴∴∴ I = 2 2

–4 9

5 5++ 4 +9

dxx x

∫

= ( )2 22

–4 9+

5 5+ 4 + 3

dx dx

x x∫ ∫

=–1 –1× ×× ×× ×× ×

–4 1 9 1tan + tan +

5 2 2 5 3 3

x xc

∴∴∴∴ I =–1 –1–2 3

tan + tan +5 2 5 3

x xc

Q-7) (((( )))) (((( ))))

2

2 2

2 – 3

+1 – 4

xdx

x x∫∫∫∫

Ans. Let I = ( ) ( )

2

2 2

2 – 3

+ 4 – 9

xdx

x x∫ ... (i)

Let x2 = t

∴∴∴∴ ( ) ( )

2

2 2

2 – 3

+1 – 4

x

x x = ( ) ( )

2 – 3

+1 – 4

t

t t

= ++1 – 4

A B

t t

∴∴∴∴ A == –1t

2 – 3

– 4

t

t

=–2 – 3

= +1–1 – 4

∴∴∴∴ B ==4t

2 – 3

+1

t

t

=( )2 4 – 3 5

= =14+1 5

∴∴∴∴ ( ) ( )

2

2 2

2 – 3

+1 – 4

x

x x =1 1+

+1 – 4t t

= 2 2

1 1+

+1 + 4x x

....(ii)


∴∴∴∴ I = 2 2

1 1+

+1 – 4dx

x x

= 2 2 2

1+

+1 – 2

dxdx

x x∫ ∫= ( )

( )–1 1 – 2

tan + log +2 2 + 2

xx c

x

∴∴∴∴ I = ( )–1 1 – 2tan + log +

4 +2

xx c

x



Q-8)(((( )))) (((( ))))2 2+1 – 4

x dx

x x∫∫∫∫

Ans. Let I = ( ) ( )2 2+1 – 4

x dx

x x∫Put x2 = t

∴∴∴∴ 2x dx = dt

∴∴∴∴ x dx = 2

dt

∴∴∴∴ I =( ) ( )

2

+1 – 4

dt

t t

∫

= ( ) ( )

1

2 +1 – 4

dt

t t∫ ... (i)

Let ( ) ( )

1

+1 – 4t t∫= +

+1 – 4

A B

t t

∴∴∴∴ A == –1t

1

– 4t

=1–5

∴∴∴∴ B ==4t

1

+1t

=1

5

∴∴∴∴ ( ) ( )1

+1 – 4t t

=

–1 –1

5 5++1 – 4t t

....(ii)


∴∴∴∴ I =

–1 11 5 5+2 +1 – 4

dtt t

∫

=1 1 1– log +1 + logg – 4 +

2 5 5t t c

=2 21

log – 4 – log +1 +10

x x c

Q-9) (((( )))) (((( ))))2– 1 +1

xdx

x x∫∫∫∫

Ans. Let I = ( ) ( )2–1 +1

xdx

x x∫ .... (i)

Let ( ) ( )2–1 +1

x

x x

=( ) ( ) ( )

( ) ( )

2

2

+1 + + –1

–1 +1

A x Bx c x

x x

∴∴∴∴ x = A(x2 + 1) + (Bx + c)(x – 1)

Put x = 1

∴∴∴∴ 2A = 1

∴∴∴∴ A = 1

2

Put x = 0

∴∴∴∴ A – C = 0

∴∴∴∴ A = C ⇒⇒⇒⇒ C = 1

2

Put x = –1

∴∴∴∴ 2A + (C –B) (–2) = –1

∴∴∴∴ 1 – 21–

2B

= –1

∴∴∴∴ 1 – 1 – 2B = –1 ⇒⇒⇒⇒ 2B = 1

∴∴∴∴ B = 1

2

∴∴∴∴ ( ) ( )2–1 +1

x

x x

=2

1 11 +2 22 +

–1 +1

x

x x

....(ii)


I = 2

1 11 +2 22 +

–1 +1

x

dxx x

∫

= 2 2 2

1 1 2 1+ +

2 –1 22 +1 +1

dx x dxdx

x x x x∫ ∫ ∫=

( )

2

–1

1 1log –1 + log –1

2 4

1+ tan +2

x x

x c



Q-10)sin sin2

dx

x x++++∫∫∫∫Ans. Let I =

sin + sin2

dx

x x∫I =

sin +2sin cos

dx

x x x∫I = ( )sin 1+ 2cos

dx

x x∫

I = ( )2

sin

sin 1+ 2cos

xdx

x x∫

I = ( ) ( )2

sin

1 – cos 1+2cos

x dx

x x∫Put t = cos x ⇒⇒⇒⇒ – sin x dx = dt

I = ( ) ( )2

–

1 – 1+ 2

dt

t t∫

I =( ) ( ) ( )

–1+ 1 – 1+ 2

dt

t t t∫Let + +

1+ 1 – 1+ 2

A B C

t t t

= ( ) ( ) ( )

1

1+ 1 – 1+ 2t t t

A(1 – t)(1 + 2t) + B(1 + 2t)(1 + t)

+ C(1 + t)(1 – t) = 1

Put t = –1⇒⇒⇒⇒ A(1 + 1)(1 – 2) = 1

⇒⇒⇒⇒ A = –1

2

Put t = –1⇒⇒⇒⇒ B(1 + 2)(1 + 1) = 1

⇒⇒⇒⇒ B = 1

6

Put t = –1

2 ⇒⇒⇒⇒

1 11 – 1+

2 2C

= 1

⇒⇒⇒⇒ 1 3

2 2C

= 1 ⇒⇒⇒⇒ C = +4

3

∴∴∴∴ ( ) ( ) ( )

1

1+ 1 – 1+ 2t t t

=

–1 1 +4

2 6 3+ +1+ 1 – 1+2t t t

∴∴∴∴ I =

–1 1 4

2 6 3– + +

1+ 1 – 1+2dt

t t t

∫

I =

–1 1 4

2 6 3– + +

1+ 1 – 1+2dt

t t t

∫

I =log 1 –1 1

+ log 1+ –2 6 –1

tt

××××4 1

– log 1+ 2 +3 2

t c

=1 1

+ log 1+ cos + log 1 – cos2 6

x x

2– log 1+ 2cos +3

x c

Q-11)(((( ))))

2

2 1

– 4

xdx

x

++++

∫∫∫∫

Ans. Let ( )

2

2 + 7

– 4

x

x = ( ) ( )

2+

– 4 – 4

A B

x x

=( )

( )2

– 4 +

– 4

A x B

x

2x + 7 = A (x – 4)2 + B = Ax + (–4A + B)

On comparing the co-efficient

A = 2 and –4A + B = 7

∴∴∴∴ B = 7 + 4A = 7 + 8 =15

∴∴∴∴( )

2

2 + 7

– 4

x

x = ( ) ( )

2

2 15+

– 4 – 4x x

I =( )

2

152 +15

– 4 – 4

dx

x x∫ ∫

= 2 log |x – 4| + 15 + ( )

( )

–2+1– 4

–2+1

x+ c

I = 2 log |x – 4| – ( )

15

– 4x+ c

Q-12)(((( ))))

2

3

1

–1

x xdx

x

+ ++ ++ ++ +

∫∫∫∫

Ans. Let ( )

2

3

+ +1

–1

x x

x =

( ) ( ) ( )2 3

+ +–1 –1 –1

A B C

x x x

x2 + x + 1 = A (x – 1)2 + B (x –1)+ C ...(i)

Comparing the co-efficients we get

1 = A ∴ ∴ ∴ ∴ A = 1

Put x = 1 in (i), we get

∴∴∴∴ 3 = C ⇒⇒⇒⇒ C = 3



Q-13)(((( )))) (((( ))))

2

3 – 2

1 3

xdx

x x+ ++ ++ ++ +∫∫∫∫Ans. I =

( ) ( )2

3 – 2

+1 + 3

xdx

x x∫

( ) ( )2

3 – 2

+1 +3

x

x x=

( ) ( ) ( )2+ +

+1 +3+1

A B C

x xx

( ) ( )2

3 – 2

+1 +3

x

x x

= ( ) ( ) ( )

( ) ( )

2

2

+3 +1 +1

+1 +3

x A x + B C x

x x

+

∴∴∴∴ 3x – 2 = (x + 3) [A (x + 1) + B] + C (x + 1)2

when x = – 1

3 (–1) – 2 = (–1 + 3) [A (0) + B] + C (–2)2

–11 = 4C ∴∴∴∴ C = –11

4

when x = 0, 3 (0) – 2 = 3 [A (0 + 1) + B]

+ C [0 +1]2

– 2 = 3A + 3B + C

–2 = 3A + 5 11

3 – –2 4

3A =15 11

–2+ +2 4

= 33

4

∴∴∴∴ A = 11

4

∴∴∴∴ I = ( )

2

11/4 5/2 11/4– –

+1 + 3+1dx

x xx

∫

= ( )

2

11 5 11– –

4 +1 2 4 + 3+1

dx dx dx

x xx∫ ∫ ∫

I = 11 5 1

log +1 – –4 2 +1

xx

11– log + 3 +4

x c

Q-14)(((( )))) (((( ))))2

8

2 4dx

x x+ ++ ++ ++ +∫∫∫∫

Ans. I = ( ) ( )2

8

+ 2 + 4x x = 2

++

+ 2 + 4

A Bx C

x x

8 = A (x2 + 4) + (Bx + c) (x + 2) ...(i)

8 = (A + B)x2 + (2B + c)x (4A + 2C) ...(ii)

Putting x = – 2 in equation (i) we get

8 = 8A ⇒ A = 1

Comparing co-efficient of x2 and constant

term on both sides, we get

A + B = 0 and 4A + 2C = 8

B = – 1 and 4 + 2C = 8 ⇒ C = 2

( ) ( )2

8

+ 2 + 4x x = 2

1 – + 2+

+ 2 + 4

x

x x

∴∴∴∴ I = 2

1 – + 2+

+2 + 4

xdx

x x

∫

= 2

1 1 2–

+ 2 2 + 4

xdx dx

x x∫ ∫ 2 2

1+2

+2dx

x∫I = log |x + 2|

2 –11 1

– log + 4 +2. tan +2 2 2

xx c

I = –1

2

+2log + tan +

2+ 4

x xc

x

Q-15)

(((( )))) (((( ))))

2

2 2

37

– 7 4

++++

++++∫∫∫∫x

dxx x

Ans. Let I = ( ) ( )

2

2 2

+ 37

– 7 + 4

xdx

x x∫

= ( ) ( )

2

2 22 2

+ 37+

– 7 + 4– 7 + 4

x A B

x xx x= ...(i)

x2 + 37 = A (x2 + 4) + A (x2 + 4) ... (ii)

Putting x2 = 7 in (ii) we get

∴∴∴∴ 7 + 37 = A (7 + 4)

44 = 11 A

A = 4

Putting x2 = – 4 in (ii) we get

–4 + 37 = B (–4 – 7)

∴∴∴∴ Put x = 0 in (i), we get

1 = A – B + C ; 1 = 1 – B + 3

∴∴∴∴ B = 3

∴∴∴∴( )

2

3

+ +1

–1

x x

x=

( ) ( ) ( )2 3

+ +–1 –1 –1

A B C

x x x

I = ( ) ( )–2 –31

+3. –1 +3 –1–1

x x dxx

∫

I = log |x – 1| – ( ) ( )

2

3 3–

–1 2 –1x x

+ c

= ( )

11 +1 5log + +

4 + 3 2 +1

xc

x x



33 = –11 B

B = –3

∴∴∴∴ From (i) we get

( ) ( )

2

2 22 2

+ 37 4 –3= +

– 7 + 4– 7 + 4

x

x xx x

I = 2 2

1 14 – 3 =

– 7 + 4dx dx

x x∫ ∫=

( )2 2 2

2

1 14 – 3 =

+ 2– 7dx dx

xx∫ ∫

= –12 – 7 3log – tan +

2 27 + 7

x xc

x

GROUP (N) – CLASS WORK PROBLEMS

Q-1)5sin xdx∫∫∫∫

Ans. I =5sin x dx∫

=4sin sinx x dx∫

= ( )4

21 – cos .sinx x dx∫cos x = t

∴∴∴∴ –sin x dx = dt

∴∴∴∴ sin x dx = dt

∴∴∴∴ I = ( ) ( )2

21 – –t dt∫= ( )

22 41 – 2 +t t dt∫

=3 52 1

– – + +3 5

t t t c

=3 52 1

cos – cos + cos +3 5

x x x c

Q-2)4cos x dx∫∫∫∫

Ans. I =4cos x dx∫

= ( )2

2cos x dx∫=

21+ cos2

2

xdx

∫

=21

1+2cos2 + cos 24

x x dx ∫=

1 1+cos41+2cos2 +

4 2

xx dx

∫

=1 1 sin4

+ sin2 + + +4 2 4

xx x x c

∫

Q-3)5tan x dx∫∫∫∫

Ans. I =5tan x dx∫

=3 2tan tanx x dx∫

= ( )3 2tan sec –1x x dx∫=

3 2 3tan sec – tanx x dx x dx∫ ∫In the first integral, put tan x = t

∴∴∴∴ sec2 x dx = dt.

∴∴∴∴ I =3 2– tan .tant dt x x dx∫ ∫

= ( )4 2

1

1– tan sec –1 +

4t x x dx c∫

= ( )4 2

1

1tan – tan sec – tan +

4x x x x dx c∫

= 4 21

tan – tan sec4

x x x dx∫1+ tan +x dx c∫

Again, in the first integral, put tan x = t,

∴∴∴∴ sec2 x dx = dt.

∴∴∴∴ I =4

1 2

1– + log sec + +

4t x t dt x c c∫

=4 2

1 2 3

1 1– + log sec + + +

4 2t x t x c c c

=4 21 1

tan – tan + log sec +4 2

x x x c

Q-4)6cot x dx∫∫∫∫

Ans. I =6cot x dx∫

I = ( )4 2cot cos –1x ec x dx∫



Q-3)4sin x dx∫∫∫∫

Ans. Let I =4sin x dx∫

= ( )2

2sin x dx∫=

21 – cos2

2

xdx

∫

=21

1 – 2cos2 + cos 24

x x dx ∫=

1 1+ cos 41 – 2cos2 +

4 2

xx dx

∫

=1 2sin2 1 sin4

– + + +4 2 2 8

x xx x c

=1 sin43 – 2sin2 + +

8 4

xx x c

Q-5)6sec x dx∫∫∫∫

Ans. I =6sec x dx∫

I = ( )2

4 2 2 2sec .sec sec secx x dx x x dx∫ ∫I = ( )

22 21+ tan .secx x dx∫

Put tan x = t


∴∴∴∴ I = ( )2

21+ t dt∫= ( )2 41+2 +t t dt∫=

3 52 1+ + +3 5

t t t c

=3 52 1

tan + tan + tan +3 5

x x x c


Ans. I =2 2cos .cosec x ec x dx∫

I = ( )2 21+cot .cosx ec x dx∫Put cot x = t ⇒⇒⇒⇒ cos ec x dx = –dt

∴∴∴∴ I = ( ) ( )21+ –t dt∫=

3

– + +3

tt c

∴∴∴∴ I = – cot x – 1

3 cot3x + c

GROUP (N) – HOME WORK PROBLEMS

Q-1)3sin x dx∫∫∫∫

Ans. I =3sin x dx∫ =

2sin .sinx x dx∫= ( )21– cos .sinx x dx∫

Put cos x = t

∴∴∴∴ sin x dx = –dt

∴∴∴∴ I = ( ) ( )21– –t dt∫ =

3

– +3

tt c

=

3cos– cos +

3

xx c

Q-2)5cos x dx∫∫∫∫

Ans. I =5cos x dx∫ = ( )

221 – cos cosx x dx∫

Put sin x = t ⇒⇒⇒⇒ cos x dx = dt

∴∴∴∴ I = ( )2

21 – t dt∫= ( )2 41– 2 +t t dt∫

I =3 52 1

sin – sin + sin +3 5

x x x c

I = ( )

4 2

2 2

cot cos –

cot cos –1

x ec xdx

x ec x

∫

I = ( )4 2 2

2

cot – cot cos

+ cot

x x ec x dx

x dx

∫∫

Put cot x = t in first integral, cosec2x dx = – dt

I = ( ) ( )2 4 2– + cos –1t t dt ec x dx∫ ∫I =

3 5cot cot– – cot – +

3 5

x xx x c




Ans. I =5cot x dx∫ =

3 2cot cotx x dx∫I = ( )3 2cot cos –1x ec x dx∫I = ( )3 2 2cot cos – cot cotx ec x x x dx∫

Q-4)4tan x dx∫∫∫∫

Ans. I =4tan x dx∫ =

2 2tan .tanx x dx∫= ( )2 2sec –1 .tanx x dx∫= ( )2 2 2sec .tan – sec –1x x dx x dx∫ ∫

In first integral,

put tan x = t ⇒⇒⇒⇒ sec2 x dx = dt.

∴∴∴∴ I =2 2– sec +t dt x dx dx∫ ∫ ∫

=

3

– tan + +3

tx x c

=

3tan– tan + +

3

xx x c


Ans. I =4cot x dx∫ =

2 2cot .cotx x dx∫= ( )2 2cos –1 cotec x x dx∫=

( )

2 2

2

cot .cos

– cos –1

x ec x dx

ec x dx

∫∫

Put cot x = t in first integral,

∴∴∴∴ –cosec2x dx = – dt

⇒⇒⇒⇒ cosec2x dx = – dt

∴∴∴∴ I = ( )2– – – cot – +t dt x x c∫=

3

– + cot + +3

tx x c

=31

– cot + cot + +3

x x x c

Q-7)4sec x dx∫∫∫∫

Ans. Let I =4sec x dx∫ = ( )2 2sin .sinx x dx∫

= ( )2 21+ tan .secx x dx∫Put tan x = t

∴∴∴∴ sin2 x dx = dt

∴∴∴∴ I = ( )21+ t dt∫ =

3

+ +3

tt c

∴∴∴∴ I =31

tan + tan + 33

x x


Ans. I =6cosec x dx∫

= ( )2

2 2cos cosec x ec x dx∫I = ( )

22 21+ cot cosx ec x dx∫

Put cot x = t ⇒⇒⇒⇒ cosec2x dx = – dt

I = ( )2

21+ t dt∫ = ( )2 4– 1+2 +t t dt∫I =

3 32 1– cot + cot + cot +

3 5x x x c

I = ( )

3 2

2

cot cos

– cot cos –1

x ec xdx

ec x

∫

I = ( )3 2cot – cot cos

+ cot

x x ec x dx

x dx

∫∫

Put cot x = t in first integral cos ec2x dx =–dt

I = ( )3 – + log sin +t t dt x c∫I =

2 4cot cot– + log sin +

2 4

x xx c



BASIC ASSIGNMENTS (BA)

BA – 1

Q-1)1

1+ sin∫∫∫∫ dxx

Ans. I =( ) ( )

1 – sin

1+ sin 1 – sin∫x

dxx x

= 2

1– sin

1 – sin∫xdx

x

= 2

1 – sin

cos∫xdx

x

= ( )2sec – tan sec∫ x x x dx

=2sec – tan sec∫ ∫x dx x x dx

I = tan x – sec x + c

Q-5)tan

sec + tan∫∫∫∫x

dxx x

Ans. I = tan

sec + tan

xdx

x x∫= ××××

tan sec – tan

sec + tan sec – tan∫x x x

dxx x x x

=

2

2 2

sec tan – tan

sec – tan∫x x x

dxx x

= ( )2sec tan – sec –1 ∫ x x x dx

=2sec tan – sec + 1∫ ∫ ∫x x dx x dx dx

= sec x – tan x + x + c

Q-4) 2 2

cos2

sin .cos∫∫∫∫x

dxx x

Ans. =

2 2

2 2

cos – sin

sin .cos∫x x

dxx x

= 2 2

1 1–

sin cos

∫ dx

x x

=2 2cosec – sec∫ ∫x dx x dx

= – cot x – tan x + c

Q-2)

2

2

– 3

+1∫∫∫∫x

dxx

Ans. I =

2

2

– 3

+1∫x

dxx

=

2

2

+1 – 4

+1∫x

dxx

=

2

2 2

+1 4–

+1 +1

∫x

dxx x

= 2

1– 4

+1∫ ∫dx dxx

I = x – 4 tan–1 x + c

Q-3) 0cos2∫∫∫∫ x dx

Ans. I =2

cos180

π

∫ x dx

=ππππ

ππππ

180 2sin +

2 180∫x

c

I =

0

ππππ

90 sin2x + c

BA – 2

Q-1)1

++++∫∫∫∫ dxx x

Ans. Let I =1

+dx

x x∫=

( )1

+1dx

x x∫

I = ( )2

2 +1dx

x x∫f (x) = ( )+1x

f ′′′′(x) =1

2 x

∴∴∴∴ I = 2 log +1 +x c



Q-2) (((( ))))1

2 + log xx x∫∫∫∫

Ans. I =1

2 + log∫ dxx x

=1

2 + log∫ dxx x x

I = ( )

1

2+ log∫ dxx x

Put log x = t ∴∴∴∴1

xdx = dt

I =1

2+∫ dtt

= log 2+ +t c

I = log 2+ log +x c

Q-3)1 – tan

1+ tan

xdx

x∫∫∫∫Ans. I =

1 – tan

1+ tan

xdx

x∫

=cos – sin

sin + cos∫x x

dxx x

f (x) =sin x + cos x f ′′′′ (x) = cos x – sin x

I =( )

( )

f xdx

f x

′

∫= log cos +sin +x x c

Q-4)(((( ))))

(((( ))))

sin –

sin +

x adx

x a∫∫∫∫Ans. I =

( )

( )

sin –

sin

x adx

x a+∫

=( )

( )

sin + – 2

sin +

x a adx

x a∫

=( ) ( )

( )

sin + cos2 – cos + sin2

sin +

x a a x a adx

x a∫

= ( )cos2 – cot + sin2adx x a adx∫ ∫∴∴∴∴ I = x cos 2a – sin 2a log |sin (x + a) | + c

Q-5)–3

1

+dx

x x∫∫∫∫Ans. I =

1

+dx

x x∫ –3

=+1

xdx

x∫3

4

Put x4 = t 4x3 dx = dt

I =1 1

4 1+dtt∫

=1log 1+ +

4t c

=1log 1+ +

4x c4

BA – 3

Q-1)2

1

7 – 6 –dx

x x∫∫∫∫Ans. I =

1

7 – 6 –dx

x x∫ 2

=( )( )2

1

2 – + 2 3 + 9dx

x x∫

=

( ) ( )2 2

1

2 – + 3

dx

x∫

=–1 + 3

sin +2

xc

Q-2)3

4 – 5sindx

x∫∫∫∫Ans. I =

3

4 – 5sindx

x∫

Put tan2

xt=

x = 2 tan t

dx =2

1+

dt

t2

∴∴∴∴ sin x =2

1+

t

t2

=2

2

3 22

2 1+4 – 5

1+

dt

t t

t

∫



= 2

32

4+ 4 –10dt

t t∫= 2

32

4 –10 + 4dt

t t∫=

2

6 1

104– +14

dt

t t∫

=2

3 1

52– +1

2

dt

t t

∫

=2

3 1

5 25 92– 2 + –

4 16 16

dt

t t

∫

=2 2

3 1

2 5 3– –4 4

dt

t

∫

=

5 3– –

3 1 4 4log +5 332 – +24 44

tc

t

I =

tan – 42log +

2tan –12

x

cx

Q-3) –

1

3 +2x xdx

e e∫∫∫∫Ans. I =

1

3 +2dx

e e∫ –x x

=3 +2

edx

e∫x

2x

Put ex = t ∴∴∴∴ ex dx = dt

I =2

1 1

23+3

dt

t∫

=2

1 1

23+3

dt

t∫

=1 1

3 2+

3t

∫ 2

2

= –11 1tan +

3 2 2

3 3

tc

I =–1 x1 3

tan +26e c

Q-4) 2

1

2 – 3cosdx

x∫∫∫∫Ans. I =

1

2 – 3cosdx

x∫ 2

Divide by cos2 x

I =sec

2sec – 3

xdx

x∫2

2

= ( )sec

2 1+ tan – 3

xdx

x∫2

2

=sec

2+2tan – 3

xdx

x∫2

2

=sec

2tan –1

xdx

x∫2

2

Put tan x = t


I =1

2 –1dt

t∫ 2

=

1 1

12–2

dt

t∫ 2

=

1 1

2 1–

2

dt

t

∫2

=

1–

1 1 2log +

12 1 +2 –22

t

c

t

∫

I =1 2 tan –1

log +2 2 2 tan +1

xc

x

Q-5) 2

sin

9 – sin

xdx

x∫∫∫∫Ans. I =

sin

9 – sin

xdx

x∫ 2

=sin

9 –1+cos

xdx

x∫ 2

I =sin

8+cos

xdx

x∫ 2

Put cos x = t –sin x dx = dt

I =–1

8+dt

t∫ 2



=( )

1–1

+ 2+ 2

dt

t∫ 2

2

= –1 log + +8 +t t c

2

I = – log cos + 8+ cos +x x c2

BA – 4

Q-1)

2

4 2

+14

+ – 2

xdx


2

4 2

+14

+ – 2

xdx

x x∫

= ( ) ( )

2

2 2

+14

+ 2 –1

xdx

x x∫ .... (i)

( ) ( )

2

2 2

+14

+ 2 –1

x

x x= ( ) ( )

+14

+ 2 –1

t

t t

= ++ 2 –1

A B

t t

Where t = x2 and

∴∴∴∴ A = =–2t

+14= –4

–1

t

t

∴∴∴∴ B = =1t

+14= 5

+ 2

t

t

∴∴∴∴ ( ) ( )

2

2 2

+14

+ 2 –1

x

x x =–4 5

++ 2 –1t t

= 2 2

–4 5+

+ 2 –1x x ... (ii)


∴∴∴∴ I = 2 2

–14 5+

+2 –1dx

x x

∫

=

( ) ( )2 222

1– 4 +5

– 1+ 2

dxdx

xx∫ ∫

= –1– 4 5 –1tan + log +

2 +12 2

x xc

x

Q-2)2

5 +2

– 3 + 2

xdx

x x∫∫∫∫

Ans. Let I =

2

2

+2

– 3 +2

xdx

x x∫

= ( ) ( )

5 + 2

– 2 –1

xdx

x x∫ .... (i)

Let ( ) ( )

5 + 2

– 2 –1

x

x x

= +– 2 –1

A B

x x

Where

∴∴∴∴ A = =2x

5 + 2=12

–1

x

x

∴∴∴∴ B = 5 +2

= –7– 2

x

x

=1x

∴∴∴∴ ( ) ( )

5 + 2

– 2 –1

x

x x =12 7

–– 2 –1x x

∴∴∴∴ I =12 7

–– 2 –1

dxx x

∫

= 12 log|x – 2| – 7 log |x – 1| + c

Q-3)1+ cos

xdx

x∫∫∫∫

Ans. Let I =1+cos

xdx

x∫=

22cos2

xdx

x

∫

=21

.sec2 2

xx dx

∫

= ( )2

tan1 2sec – 1

12 2

2

xx

x dx dx

∫ ∫

=

tan1 2

. – 2 tan12 2

2

x

xx dx

∫



=

2log sec21

2 .tan – +12 2

2

x

xx c

=

2log sec21

2 .tan – +12 2

2

x

xx c

= .tan – 2log sec +2 2

x xx c

Q-4)(((( ))))

2+1

xxe

dx

x∫∫∫∫

Ans. I =( )

2

x

+1

xedx

x∫

=( )

2

x +1 –1

+1

xe dx

x

∫

=( ) ( )

2 2

x +1 1–

+1 +1

xe dx

x x

∫

I =( ) ( )

2

x 1 1–

+1 +1e dx

x x

∫

f(x) = ( )1

+1x

f ′′′′(x) = ( )

2

–1

+1x

I = 1

. ++1

e cx

x

I = ++1

ec

x

x

Q-5) 29 – 16x dx∫∫∫∫

Ans. Let I =29 –16x dx∫

I =2 16

3 –9

x dx∫

I =

2

2

16–

2 9

3 +16169– log + –

2 9

xx

c

x x

I = 2 23 16 8 16– – log + – +

2 9 3 9

xx x x c

BA – 5

Q-1) sin x dx∫∫∫∫Ans. Let I = sin x dx∫

Put x = t

∴∴∴∴1

x xdx = dt

∴∴∴∴ dx = 2t dt

∴∴∴∴ I = 2 sint t dt∫= 2 sint t dt∫= 2 sin – sin

dtt t dt t dt dt

dt

∫ ∫ ∫ ∫

= ( )2 – cot + cos +t t t dt c ∫

= [ ]2 – cos +sin +t t t c

I = –2 cos + 2sin +x x x c

Q-2) (((( ))))log

1+ log

xdx


( )

log

1+ log

xdx

x∫Put log x = t

∴∴∴∴ x = et


∴∴∴∴ I = ( )2

t

1+

te dt

t∫

=( )

( )2

t+1 –1

1+

te dt

t

∫



=( ) ( )

2 2

t +1 1–

1+ 1+

te dt

t t

∫

=( )

1 1–

1+ 1+e dt

t t

∫ 2

t

=( )

2

t 1 1–

1+ 1+e dt

t t

∫

Now, ( )2

1 –1=

1+ 1+

d

dt t t

∴∴∴∴ I =t 1 1

+1+ 1+

de dt

t dt t

∫

=t 1. +1+

e ct

∴∴∴∴ I = +1+ log

xc

x

Q-2) (((( ))))log

1 log

xdx

x++++∫∫∫∫

Ans. Let I = ( )

log

1+ log

xdx

x∫Put log x = t

∴∴∴∴ x = et


∴∴∴∴ I =( )

2

t

1+

te dt

t∫

=( )

( )2

t+1 –1

1+

te dt

t

∫

=( ) ( )

2 2

t +1 1–

1+ 1+

te dt

t t

∫

=( )

2

t +1 1–

1+ 1+

te dt

t t

∫

=( )

2

t 1 1–

1+ 1+e dt

t t

∫Now,

( )2

1 –1=

1+ 1+

d

dt t t

∴∴∴∴ I =t 1 1

+1+ 1+

de dt

t dt t

∫

=t 1. +1+

e ct

∴∴∴∴ I = +1+ log

xc

x

Q-3)3 + 4 sin

dx


3+ 4sin

dx

x∫Put tan

2

x = t sin x = 2

2

1+

t

t

dx = 2

2

1+

dt

t

I =( )2

2

1 2

2 1+3 + 41+

dt

t t

t

∫

= 23+3 +8

xdt

t t∫=

2

2 1

83+ +13

dtt

t∫

=2

2 1

4 16 73+ 2 + –

3 9 9t t

∫

= 22

2 1

3 4 7+ –3 3

t

∫

=2 1 + 4/3 – 7 /3

log +3 + 4/3 + 7 /372

3

tc

t

I =3tan + 4 – 7

1 2log +7 + 4/3+ 7 /3

x

ct

Q-4)–1tan xdx∫∫∫∫

Ans. I =–1tan x dx∫

=–11.tan x dx∫



=–1 –1tan – tan

dx dx x dx dx

dx

∫ ∫ ∫

=–1

2

1tan –

1+x x x dx

x∫=

–1

2

1 2tan –

2 1+

xx x dx

x∫I = ( )–1 21

tan – log 1 +2

x x x c+

Q-5)sin + sin2

dx

x x∫∫∫∫Ans. Let

I =sin +sin2

dx

x x∫=

sin +2sin cos

dx

x x x∫= ( )sin 1+ 2cos

dx

x x∫

= ( )

sin

sin + 1+2cos

x dx

x x∫ 2

= ( ) ( ) ( )sin

1 – cos 1+ cos 1+2cos

x dx

x x x∫Put cos x = t

∴∴∴∴ –sin x dx = dt

∴∴∴∴ sin x dx = –dt

∴∴∴∴ I = ( ) ( ) ( )–

1– 1+ 1+2

dt

t t t∫

=( ) ( ) ( )

–1– 1+ 1+2

dt

t t t∫

Let ( ) ( ) ( )1

1– 1+ 1+2t t t

= + +1 – 1+ 1+2

A B C

t t t

∴∴∴∴ 1 = A(1 + t)(1 + 2t) + B(1 – t) (1 + 2t) +

C(1 – t) (1 + t)

Putting 1 – t = 0,

i.e., t = 1, we get,

1 = A(2)(3) + B(0)(3) + C(0)(2)

∴∴∴∴ A = 1

6

Putting 1 + t = 0, t = –1, we get,

1 = A(0)(–1) + B(2)(–1) + C(2)(0)

∴∴∴∴ B = 1

–2

Putting 1 + 2t = 0, i.e., t = 1

–2, we get,

1 = ( ) ( )3 1

0 02 2

A B C

+ +

∴∴∴∴ C = 4

3

∴∴∴∴ ( ) ( ) ( )1

1– 1+ 1+2t t t

=

1 –1 4

6 2 3+ +

1– 1+ 1+2t t t

∴∴∴∴ I =

1 –1 4

6 2 3– + +

1– 1+ 1+2dt

t t t

∫

=1 1 1 1 4 1

– + –6 1 – 2 1+ 3 1+2

dt dt dtt t t∫ ∫ ∫

=log 1 –1

– .6 –1

t

log 1+21 4+ log 1+ – . +2 3 2

tt c

=1 1log 1 – cos + log 1+ cos

6 2x x

2– log 1+2cos + .3

x c



Q-2) secn x . tan x

Ans. I = sec .tanx x dx∫ n

= sec .sec tanx x x∫ –1n

Put sec x = y

∴∴∴∴ sec x tan x dx = dy

∴∴∴∴ I =1

= + = +–1+1

yy dy c y c

n n∫–1+1

–1n

n n

∴∴∴∴ I =1

nsecn x + c

Q-3)cos2 – cos2

cos cos

x

x

αααα

αααα⋅⋅⋅⋅

Ans. I =cos2 – cos2

cos .cos

xdx

x

α

α∫

=( ) ( )2cos –1 – 2cos –1

cos cos

xdx

x

α

α⋅∫2 2

=2cos –1 – 2cos +1

cos cos

xdx

x

α

α⋅∫2 2

=2cos 2cos

–cos cos cos cos

xdx

x x

α

α α

⋅ ⋅ ∫2 2

=cos cos

2 – 2cos cos cos cos

xdx dx

x x

α

α α⋅ ⋅∫ ∫2 2

= 2sec cos – 2cos secx dx x dxα∫ ∫= 2sec α sin x – 2 cos α sec + tan +x x c

∴∴∴∴ I =2

cosαsin x – 2 cos α + log

sec + tan +x x c

Q-4) (((( )))) (((( ))))2 3 7 3.cos sinx x x dx∫∫∫∫

Ans. I = ( ) ( )cos sinx x x dx∫ 2 3 7 3

Put sin x3 = y

∴∴∴∴ 3x2 cos x3 dx = dy

∴∴∴∴ x2 cos x3 dx = 1

3 dy

∴∴∴∴ I =

+

1 1 1= = +

73 3 3+1

2

yy dy y dy c∫ ∫

717 2

7 2

=1 2

+3 9

y c9

2× ×× ×× ×× ×

∴∴∴∴ I = ( )2sin +

27x c

93 2

AA–2

Q-1) (((( ))))–1cot sec3 tan3 dθ θ θθ θ θθ θ θθ θ θ++++∫∫∫∫Ans.

1 sin3 1+ sin3sec3 + tan3 = + =

cos3 cos3 cos3

θ θθ θ

θ θ θ

=

1 – cos2+3

sin2+3

π

θ

π

θ

ADVANCED ASSIGNMENTS (AA) :

AA–1

Q-1) sin2

sin – sin + +6 6

x

x x cπ ππ ππ ππ π

∫∫∫∫

Ans. I = ( ) ( )

sin2

sin – 6 sin + 6

xdx

x xπ π∫

=

sin – + +6 6

sin – + +6 6

x x

x x

π π

π π

∫

=

sin – cos +6 6

+ cos – sin +6 6

sin – sin +6 6

x x

x x

dx

x x

π π

π π

π π

∫

= cot + + cot –6 6

x x dxπ π

∫

= log sin + + log sin – +6 6

x x cπ π



=

32sin +

4 2

3 32sin + cos +

4 2 4 2

π θ

π θ π θ

2

=3

tan +4 2

π θ

∴∴∴∴ I =3

cot tan +4 2

dπ θ

θ

∫ –1

=3

cot cot – –2 4 2

dπ π θ

θ

∫ –1

=3

–4 2

dπ θ

θ ∫

=3

– +4 4

cπ θ

θ2

Q-2) sin

sin3

xdx

x∫∫∫∫

Ans. Let I =sin

sin3

xdx

x∫=

sin

3sin – 4sin

xdx

x x∫ 3

=1

3 – 4sindx

x∫ 2

Divide numerator and denominator by sin2x

I =cosec

3cosec – 4

xdx

x∫2

2

=cosec

3cot –1

xdx

x∫2

2

Put cot x = 1 ∴∴∴∴ cosec2x dx = – dt

I =1

–3 –1

dtt∫ 2

=

1 1–

13–3

dt

t∫ 2

=1 1

–3 1

–3

dt

t

∫ 2

2

=

1–

1 1 3– log +

13 1 +233

t

c

t

××××

=1 3 –1

– log +2 3 3 +1

tc

t

∴∴∴∴sin

sin3

xdx

x∫ = 1 3 cot –1

– log +2 3 3 cot +1

xc

x

Q-3) 2

1

cos2 + 3sindx

x x∫∫∫∫

Ans. Let I =1

cos2 +3sindx

x x∫ 2

=1

1– 2sin +3sindx

x x∫ 2 2

=1

1+sindx

x∫ 2

Dividing both numerator and denominator by

I =sec

sec + tan

x dx

x x∫2

2 2

=sec

1+ tan + tan

x dx

x x∫2

2 2

=sec

2tan +1

x dx

x∫2

2

Put tan x = t ∴∴∴∴ sec2x dx = dt

∴∴∴∴ I =1 1 1

=22 +1 1

+2

dt dtt

t

∫ ∫2 2

2

=1 1

tan +12 1

22

tc

–1××××

= ( )1tan 2 tan +

2x c–1



Q-4) 2 + sin2

1+ cos2

x xe dx

x

∫∫∫∫

Ans. Let I =2+sin2

1+cos2

xe dx

x

∫ x

=2+2sin cos

2cos

x xe dx

x

⋅ ∫ 2

x

=2 2sin cos

+2cos 2cos

x xe dx

x x

⋅ ∫ 2 2

x

= sec + tane x x dx ∫ 2x

= tan +sece x x dx ∫ 2x

∴∴∴∴2+sin2

1+cos2

xe dx

x

∫ x = ex tan x + c

Q-5) 2 – sin2

1 – cos2

x xe

x

Ans. I =2 – sin2

1– cos2

xe dx

x

∫ x

=2 – 2sin cos

2sin

x xe dx

x

∫ 2

x

=1 sin cos

–sin sin

x xe dx

x x

∫ 2 2

x

= ( )cosec – cote x x dx∫ 2x

= (–1) ( )cot – cosece x x dx∫ 2x

Let f (x ) = cos x

∴∴∴∴ f ′′′′(x ) = cosec2x

∴∴∴∴ I = ( ) ( ) ( )+ = +e f x f x dx e f x c ∫ x x

= – ex cot x + c

∴∴∴∴ I = (– 1) ex cot x + c

AA–3

Q-1) (((( ))))

2

2

log – 1

1+ log

xdx

x

∫∫∫∫

Ans. I =( )

log –1

1+ log

xdx

x

∫

2

2

Put log x = t

∴∴∴∴ x = et


∴∴∴∴ I =–1

1+

te dt

t

∫

2

2

t

=( )

– 2 +1

1+

t te dt

t

∫2

22

t

=

( )

( ) ( )

1+ 2–

1+ 1+

t te dt

t t

∫2

2 22 2

t

= ( )

1 2–

1+ 1+

te dt

t t

∫ 2 22

t

Now, 1

1+

d

dt t

2 = ( )1+d

tdt

–12

= (–1) (1 + t2)–1–1d

dx(1+t2)

= (–1) (1+ t2)–2 (0 + 2t)

= ( )

– 2

1+

t

t2

2

∴∴∴∴ I =1 1

+1+ 1+

de dt

dtt t

∫ 2 2

t

=1

+1+

e ct

2

t

=( )

1+

1+ logx c

x

∴∴∴∴ I =( )

+1+ log

xc

x2



Q-2) 1

cos + sin2dx

x x

∫∫∫∫

Ans. I =1

cos +sin2dx

x x

∫

=1

cos + 2sin cosdx

x x x

∫

= ( )

1

cos 1+ 2sindx

x x∫Put sin x = t


∴∴∴∴ dx =cos

dt

x

∴∴∴∴ I =( )cos

cos 1+2sin

dt

x

x x∫

=1+2sin

dt

x∫

= ( ) ( )1 – sin 1+ 2sin

dt

x x∫ 2

= ( ) ( ) ( )1+sin 1 – sin 1+ 2sin

dt

x x x∫

= ( ) ( ) ( )1+ 1 – 1+ 2

dt

t t t∫

Let ( ) ( ) ( )

1

1+ 1 – 1+ 2t t t = + +

1+ 1 – 1+ 2

A B c

t t t

=

( ) ( ) ( ) ( )

( ) ( )

( ) ( ) ( )

1 – 1+ 2 + 1+ 1+ 2

+ 1+ 1 –

1+ 1 – 1+ 2

A t t B t t

C t t

t t t

∴∴∴∴ 1 = A (1 – t ) (1 + 2t ) + B (1 + t ) (1 + 2t )

+ C (1 + t ) (1 – t )

Putting t = – 1, we get

1 = A (1 + 1) (1 – 2) + B (0) + C (0)

∴∴∴∴ A = –1

2

Putting t = 1, we get

1 = A(0) + B(1 + 1) (1 + 2) + C(0)

∴∴∴∴ B = 1

6

Putting t = 1

2, we get

A (0) + B (0) + C1 1

1 – 1+2 2

∴∴∴∴ C = 4

3

∴∴∴∴( ) ( ) ( )

1

1+ 1 – 1+ 2t t t=

–1 1 4

2 6 3+ +1+ 1 – 1+ 2t t t

∴∴∴∴ I =

–1 1 4

2 6 3+ +1+ 1 – 1+2

dtt t t

∫

=1 1 4

+ +2 1+ 6 1 – 3 1+2

dt dt dt

t t t∫ ∫ ∫=

log 1 –1 1log 1+ +

2 6 –1

tt

log 1+24+ +3 2

tc

∴∴∴∴ I = –1 1log 1+ sin – log 1 – sin

2 6x x

2+ log 1+ 2sin +3

x c

Q-3) (((( )))) (((( ))))2

3 – 2

+1 + 3

xdx

x x∫∫∫∫

Ans. I =( ) ( )

3 – 2

+1 + 3

xdx

x x∫ 2

( ) ( )

3 – 2

+1 +3

x

x x2 =

( ) ( ) ( )+ +

+1 +1 + 3

A B C

x x x

3x – 2 = A(x + 1) (x + 3) + B (x + 3)

+ C(x + 1)2 ...(i)

Put x = – 1 in (i), we get

– 5 = B(2) ⇒⇒⇒⇒ B = –5

2

Put x = – 3 in (1), we get

–11 = 4 C C = –11

4

Comparing co-efficient of x2

A + C = 0



A –11

4= 0 ⇒⇒⇒⇒ A =

11

4

∴∴∴∴ I =( ) ( ) ( )

–511 –11

24 4+ ++1 +3+1

dxx xx

∫ 2

= ( )

11 1 5 1 11– –

4 +1 2 4+1dx dx

x x∫ ∫ 2

( )

1

+3dx

x∫

= ( )

11 5 1log +1 +

4 2 +1x

x

11– log + 3 +4

x c

=( )

11 +1 5log + +

4 + 3 2 +1

xc

x x

Q-4)

2

3 2

5 + 20 +

+ 2 +

x x cdx

x x x∫∫∫∫

Ans. Let I =5 + 20 + 6

+ 2 +

x xdx

x x x∫2

3 2

= ( )5 + 20 +6

+ 2 +1

x xdx

x x x∫2

2

=( )

5 + 20 + 6

+1

x xdx

x x∫2

2

Let ( )

5 +20 +6

+1

x x

x x

2

2 = ( )

+ ++1 +1

A B C

x x x2

∴∴∴∴ 5x2 + 20x + 6 = A (x + 1)2 + Bx (x + 1) + Cx

Put x = 0, we get,

0 + 0 + 6 = A (1) + B (0)(1) + C (0) ∴∴∴∴ A = 6

Put x + 1 = 0, i.e., x = – 1, we get,

5(1) + 20(– 1) + 6 = A (0) + B (–1) (0) + C (–1)

∴∴∴∴ – 9 = – C ∴∴∴∴ C = 9

Put x = 1, we get,

5(1) + 20(1) + 6 = A (4) + B (1)(2) + C (1)

But A = 6 and C = 9

∴∴∴∴ 31 = 24 + 2B + 9 ∴ B = – 1

∴∴∴∴( ) ( )

5 +20 +6 6 1 9= – +

+1+1 +1

x x

x xx x x

2

2 2

∴ I =( )

6 1 9– +

+1 +1dx

x x x

∫ 2

= ( )1 1

6 – +9 +1+1

dx dx x dxx x∫ ∫ ∫

–2

=( )+1

6 log – log +1 + 9 +–1

xx x c⋅

–1

=9

6 log – log +1 – ++1

x x cx

Q-5) (((( )))) (((( ))))tan – tan + .tan2x x xdxα αα αα αα α∫∫∫∫

Ans. Let I = ( ) ( )tan – tan + .tan2x x x dxα α∫Let tan 2x = tan [(x + α) + (x – α)

=( ) ( )

( ) ( )

tan + + tan –

1 – tan + tan –

x x

x x

α α

α α⋅

∴∴∴∴ tan 2x [1 – tan (x + α) tan (x – α)]

= tan (x + α) + tan (x – α)

∴∴∴∴ tan 2x – tan 2x tan (x + α) tan (x – α)

= tan (x + α) + tan (x – α)

∴∴∴∴ tan 2x – tan (x + α) – tan (x – α)

= tan 2x . tan (x + α) . tan (x – α)

∴∴∴∴ I =( )

( )

tan2 – tan +

– tan –

x xdx

x

α

α

∫

= ( )1log sec2 – log sec +

2x x α

( )– log sec – +x cα

AA–4

Q-1)

–1 –1

–1 –1

sin – cos

sin + cos

x xdx

x x∫∫∫∫

Ans. I =sin – cos

sin + cos

x xdx

x x∫–1 –1

–1 –1

=

sin – – sin2

2

x x

dx

π

π

∫

–1 –1

=2

2sin –2

x dxπ

π

∫ –1



=4

sin – 1x dx dxπ ∫ ∫–1

=4

sin – +x dx x cπ ∫ –1

=4

– +I x cπ

1

Now, I1= sin x dx∫ –1

Put x = sin2 θθθθ

dx = sin2θθθθ dθθθθ

∴∴∴∴ dx = 2 sin θθθθ . cos θθθθ dθθθθ

I = sin sin sin2 dθ θ θ ∫ –1 2

I = sin2 dθ θ θ⋅∫=

cos2 1– + cos2

2 2d

θθ θ θ∫

= ( )1 1

– 1 – 2sin + sin 1 – sin2 2

θ θ θ θ2 2

= ( )1 1

– 1 – 2 sin + 1 –2 2

x x x x⋅–1...(ii)

From (i) and (ii), we get

I = ( )4 1 1

– 1– 2 sin + – – +2 2

x x x x x cπ

–1 2

= ( )2

– – 1 – 2 sin – +x x x x x cπ

2 –1

Q-2) (((( ))))(((( ))))

2

1log log +

logx dx

x

∫∫∫∫

Ans. Let I = ( )( )

1log log +

logx dx

x

∫ 2

Put log x = t ∴∴∴∴ x = et


∴∴∴∴ I =1

log +e t e dtt

∫ 2

t t

=1 1 1

log + – +e t dtt t t

∫ 2

t

=1 1 1

log + + – +e t e dtt t t

∫ 2

t t

=1 1 1

log + – –e t dt e dtt t t

∫ ∫ 2

t t

= I1 – I

2

In I1, Put f (t ) = log t. Then f ′′′′(t ) = (1/t )

∴∴∴∴ I1= ( ) ( )+e f t f t dt′ ∫ t

= et f (t ) = et log t

In I2, Put g (t) = (1/t). Then g′′′′(t) = – (1/t2)

∴∴∴∴ I2= ( ) ( )+e g t g t dt′ ∫ t

= et g (t) = et . (1/t).

∴∴∴∴ I = et log t – e

t

t

+ c

= x log (log x) –log

x

x+ c.

Q-3) tan

1 – sin

xdx

x∫∫∫∫

Ans. Let I =tan

1 – sin

xdx

x∫

= ( )

sin

cos 1 – sin

xdx

x x∫

= ( )

sin .cos

cos 1 – sin

x xdx

x x∫ 2

=( ) ( )

sin .cos

1 – sin 1 sin

x xdx

x x+∫ 2

Put sin x = t


I =( ) ( )1 – 1+

tdt

t t∫ 2

We express

( ) ( )1 – 1+

t

t t2

=( ) ( )

+ +1 – 1+1 –

A B C

t tt2

t = A(1 – t) (1 + t) + B(1 + t) + C(1 – t)2

t = (1 + t) [A (1 – t) + B] + C (1 – t)2

...(i)

Put t = –1 in equation (i), we get

–1 = 4C ⇒⇒⇒⇒ C = –1

4



Put t = 1 in equation (i), we get

1 = 2 [0 + B] ⇒⇒⇒⇒ B = 1

2

Put t = 0 in equation (i), we get

0 = 1[A + B] + C

0 =1 1

+ –2 4

A

∴∴∴∴ A =1–4

∴∴∴∴( ) ( )1 – 1+

t

t t2

=( )

1–1 –1

24 4+ +1 – 11 –t tt +2

∴∴∴∴ I = ( )1 1 1 1 1

– + 1 – –4 1 – 2 4 1+

dt t dt dtt t∫ ∫ ∫

–2

= ( )

1 1 1log 1 – + – log 1+ +

4 2 1 – 4t t c

t

=( )

1 1 – 1log + +

4 1+ 2 1 –

tc

t t

=( )

1 1 – sin 1log + +

4 1 sin 2 1 – sin

xc

x x+

Q-4) (((( ))))log 1 cos – tan2

xx x dx

++++

∫∫∫∫

Ans. I = ( )log 1+ cos – tan2

xx x dx

∫

= ( )log 1+ cos – tan2

xx dx x dx∫ ∫ ...(i)

= ( )log 1+ cos x dx∫= ( ) ( )log 1+ cos 1x dx∫= ( ) ( )log 1+ cos

dx x dxdx∫

= ( ) ( ) ( )log 1+ cos – log 1+ cos .d

x x x x dxdx

∫

= ( )2sin cos

2 2log 1+ cos + .

2cos2

x x

x x x dxx∫ 2

= ( )log 1+ cos + .tan +2

xx x x dx c∫

From (i). we get

I = x log (1 + cos x)

+ tan + – tan2 2

x xx dx c x dx∫ ∫

∴∴∴∴ I = x log (1 + cos x) + c

Q-5) cos2

sindx

θθθθ

θθθθ∫∫∫∫

Ans. I =cos2

sind

θθ

θ∫ =1 – 2sin

sind

θθ

θ∫2

=1 – 2sin

sind

θθ

θ∫2

2=

1– 2

sindθ

θ∫ 2

= cos – 2ec dθ θ∫ 2= ( )1+ cot – 2dθ θ∫ 2

= cot –1dθ θ∫ 2=

cot –1

cot –1

θ

θ∫2

2

=cosec

– 2cot –1 cot –1

dd

θ θθ

θ θ∫ ∫2

2 2

= I1 – 2I

2...(i)

Now,

I1=

cosec

cot –1d

θθ

θ∫2

2

Put cot θθθθ = t

∴∴∴∴ – cosec2 θθθθ dθθθθ = dt

∴∴∴∴ cosec2 θθθθ dθθθθ = – dt

∴∴∴∴ I1=

––

–1 –1

dt dt

t t

=∫ ∫2 2

= – log + –1 +t t c2

1

= – log cot + cot –1 + cθ θ2 1 ...(ii)

Also, I2=

cot –1 cos–1

sin

d dθ θ

θ θ

θ

=∫ ∫2 2

2

=sin

cos – sind

θθ

θ θ∫ 2 2



I2=

sin

2cos –1d

θθ

θ∫ 2

Put cos θθθθ = u

∴∴∴∴ – sin θθθθ dθθθθ = du

∴∴∴∴ sin θθθθ dθθθθ = –du

∴∴∴∴ I2

=–

2 –1

du

u∫ 2=

––

12 –

2

du

u

∫2

=1

–2 1

–2

du

u∫ 2

=1 1

– log + – +2 2

u u c2

2

I2

=1 1

– log cos + cos – +2 2

cθ θ2

2 ...(iii)

From (i), (ii) and (iii), we get

I = – log cot + cot –1 + cθ θ21

1 1+2 log cos + cos – +

22cθ θ

2

2

– log cot + cot –1θ θ2

1+ 2 log cos + cos – +

2cθ θ2

AA–5

Q-1) cos 7 – cos8

1 2cos5

x xdx

x++++∫∫∫∫Ans. I =

cos7 – cos8

1 2cos5

x xdx

x+∫

=

7 – 8 8 – 72sin sin

2 2

51 2 1 – sin

2

x x x x

x

+

∫ 2

=

15sin .sin

2 225

3 – 4sin2

x x

dxx∫ 2

=

15 5sin .sin .sin

2 2 225 5

3sin – 4sin2 2

x x x

dxx x∫ 3

=

15 5sin .sin .sin

2 2 225

sin32

x x x

dxx

∫

=

15 5sin .sin .sin

2 2 2215

sin2

x x x

dxx∫

=5

2sin .2 2

x xdx∫

=5 5

cos – – cos +2 2 2 2

x x x xdx

∫

= ( )cos2 – cos3x x dx∫

= cos2 – cos3x dx x dx∫ ∫∴∴∴∴ I =

sin2 sin3– +

2 3

x xc

Q-2) 2 2 –

x

x x

edx

a e b e++++∫∫∫∫

Ans. Let I =+

edx

a e b e∫ 2 2 –

x

x x

=

+

edx

ba e

e

∫ 22

x

x

x

=+

edx

a e b∫ 2 2

x

2x

Put ex = t ∴∴∴∴ ex dx = dt.

∴∴∴∴ I =1

+dt

a t b∫ 2 2 2

=1 1

+

dta b

ta

∫ 2

2



=1log + + +

bt t c

a a

2

=1log + + +

be e c

a a

2

2x x

Q-3) 3 2sin2 4 cos 2

dx

x x+ ++ ++ ++ +∫∫∫∫Ans. Let I =

3+2sin2 + 4cos2

dx

x x∫Put tan x = t ∴∴∴∴ x = tan–1 t

∴∴∴∴ dx =1+

dt

t 2 and

sin 2x =2

,1+

t

t2 cos 2x = 1 –

1+

t

t

2

2

∴∴∴∴ I =1

.1+2 1 –

3 + 2 + 41+ 1+

dt

tt t

t t

∫ 22

2 2

=( ) ( )

1+.1+3 1+ + 4 + 4 1 –

t dt

tt t t∫2

22 2

=1

7 + 4 –dt

t t∫ 2

= ( )1

7 – – 4 + 4 + 4dt

t t∫ 2

=( ) ( )

1

11 – – 2

dt

t∫ 2 2

=1 11 + – 2

log2 11 11 – 2

tc

t+

+

=1 11 + tan – 2

log +2 11 11 – tan + 2

xc

x

Q-4) (((( )))){{{{ }}}}2 2

4

1 log 1 – 2 logx x x

dxx

+ ++ ++ ++ +

∫∫∫∫

Ans. Let I = ( ){ }+1 log +1 – 2logx x x

dxx∫

2 2

4

= ( )+1 1

log +1 – log . .x

x x dxx x

∫

22 2

3

=+1 +1 1

log . .x x

dxx x x

∫

2 2

2 2 3

=1 1 1

log 1+ . 1+ . dxx x x

∫ 2 2 3

Put 1

1 tx

+ =2

∴∴∴∴ –2x–3 dx = dt

∴∴∴∴1dx

x3=

–

2

dt

∴∴∴∴ I = ( )( )–

log . .2

dtt t∫

1

2

= ( )1

– log .2

t t dt∫1

2

= ( ) ( )1

– log – log2

dt t dt t t dt dt

dt

∫ ∫ ∫

1 1

2 2

= ( )1 1

– log . – .2 3/2 3/2

t tt dt

t

∫3 3

2 2

=1 2 2

– .log –2 3 3

t t t dt ∫

3 1

2 2

=1 2

– log – +2 3 3/2

tt t c

×

33 22

=1 2

– log –3 3t t c

+

3

2

=1 1 1 2

– 1+ log 1+ – +3 3

cx x

3

2

2 2

Q-5) (((( ))))

2

1 1–

log logdx

x x

∫∫∫∫

Ans. Let I = ( )

1 1–

log logdx

x x

∫ 2

Put log x = t ∴ ∴ ∴ ∴ x = et


∴∴∴∴ I =1 1– .e dt

t t

∫ 2

t

=1 1–e dt

t t

∫ 2

t



AA–6

Q-1) (((( ))))2

log 2 logxe

x x x dxx

++++ ∫∫∫∫

Ans. I = ( )log + 2loge

x x x dxx ∫

2x

= ( )2log

log +x

e x dxx

∫

2x

Put f(x ) = (log x)2

∴∴∴∴ f ′(x) = ( )logd

xdx

2 = 2 (log x). ( )log

dx

dx

=2log x

x

∴∴∴∴ I = ( ) ( )+e f x f x dx′ ∫ x

= ex. f(x) + c = ex. (log x)2 + c.

Q-2) 1 –

1

xdx

x++++∫∫∫∫

Ans. Let I =1 –

1+

xdx

x∫Put =x t ∴∴∴∴ x = t ∴∴∴∴ dx = 2t dt

I =1 –

21+

tt dt

t∫

=( ) ( )

( ) ( )

1 – 1 –2

1+ 1 –

t tt dt

t t∫

=( )1 –

21 –

tt dt

t∫2

2

=( )1 –

21 –

t tdt

t∫ 2

=–

21 –

t tdt

t∫2

2

Q-3) (((( ))))2

2

3 3

xdx

x x

++++

+ ++ ++ ++ +∫∫∫∫

Ans. I = ( )

+ 2

+3 +3

xdx

x x∫ 2

Let x + 2 = a (2x + 3) + b

x + 2 = 2ax + 3a + b

Comparing co-efficents

2a = 1 3a + b = 2 ∴∴∴∴ 3

2+ b = 2

a =1

2 b = 2 –

–3

2 ; ∴∴∴∴ b =

1

2

I =

( )1 12 + 3 +

2 23 9 3

+ 2 + +2 4 4

x

dx

x x

∫ 2

=

3 1+

2 2

3 3+ +2 4

x

dx

x

∫ 2

Let (f )t =1

t . ∴ ∴ ∴ ∴ f ′′′′(t) =

–1

t2

∴∴∴∴ I = ( ) ( )e f t f t dt′+ ∫ t

= ( )1

.e f t c e ct

+ = × +∫ t t

= +log

xc

x

= 2 –1 – 1 –

t tdt

t t

∫

2

2 2

=1 2 –1+1

2 –2 1 – 1 –

t tdt dt

t t

∫ ∫

2

2 2

= ( )1 12 –2 1 – –

2 1 –t dt

t

∫2

2

–1–

1 –

tdt

t

∫

2

2

= 2 – 1 – + cos + 1 –t t dt

2 –1 2

= 2 – 1– + cos + 1–2

tt t

2 –1 2

1+ sin2 1

t

–1

= –2 1 – +2cos + 1 – + sint t t t t2 –1 2 –1

I = –2 1 – + 2cosx x–1

+ 1– +sin +t x x c–1

I = –2 1 – + cos + 1 – +x x x x c–1



=

3 1+2 2 +3 3+ +2 4

x

dx

x

∫ 2

1

2

3 3+ +2 2

dx

x

∫ 22

=

32 +

1 2+

2 3 3+ +2 4

x

dx

x

∫ 2

1 1

2 3 3+ +2 2

dx

x

∫ 22

=1 3 3log + +2 2 4

x

2

3+

1 2+ tan3 3

2 2

x

–1

= 1log + 3 + 32

x x2

2 2 + 3+ tan +3 3

xc

–1

Q-5) (((( ))))

3

2

1

xdx

x

++++

++++∫∫∫∫

Ans. Let I =( )

+ 2

+1

xdx

x∫ 3

Let ( )

2

1

x

x

+

+3 =

( ) ( )+ +

+1 +1 +1

A B C

x x x2 3

∴∴∴∴ x + 2 = A (x + 1)2 + B(x + 1) + C ...(i)

Put x + 1 = 0, i.e., x = –1, we get,

–1 + 2 = A (0) + B (0) + C ∴∴∴∴ C = 1

Put x = 0 in (i), we get,

0 + 2 = A (1) + B (1) + C

∴∴∴∴ 2 = A + B + 1

∴∴∴∴ A + B = 1 ...(ii)

Put x = 1 in (i), we get,

1 + 2 = A(4) + B(2) + C

∴∴∴∴ 3 = 4A + 2B + 1

∴∴∴∴ 4A + 2B = 2 ∴∴∴∴ 2A + B = 1 ...(iii)

Subtracting (ii) from (iii), we get,

A = 0

∴∴∴∴ from (ii), 0 + B = 1 ∴∴∴∴ B = 1

∴ ∴ ∴ ∴ ( )

+2

+1

x

x3=

( ) ( )

0 1 1+ +

+1 +1 +1x x x2 3

=( ) ( )

1 1+

+1 +1x x2 3

∴∴∴∴ I =( ) ( )

1 1+

+1 +1dx

x x

∫ 2 3

= ( ) ( )+1 + +1x dx x dx∫ ∫–2 –3

=( ) ( )+1 +1

+ +–1 –2

x xc

–1 –2

=( )

–1 1– +

+1 2 +1c

x x2

AA–7

Q-1) 1

cos cos2dx

x x∫∫∫∫

Ans. Let I = ∫1

cos cos2dx

x x

= ∫ 2

2

sec

1 – tan

1+ tan

xdx

x

x

= ∫ 2

××××sec sec

1 – tan

x xdx

x

= ∫2

2

sec

1 – tan

xdx

x

Put tanx = t ∴∴∴∴ sec2x dx = dt

∴∴∴∴ I = ∫ 2

1

1 – tdt = sin–1 t + c

= sin–1 (tanx) + c



Q-2) (((( )))) (((( ))))–2

log log + logx x

Ans. I = ( )( )

∫ 2

1log log +

logx dx

x

Put logx = t

∴∴∴∴ x = et ∴∴∴∴ dx = etdt

∴∴∴∴ I = ∫ t

2

1log +t e dt

t

= ∫ t

2

1 1 1log + – +e t dt

t t t

= ∫ t

2

1 1 1log – + +e t dt

t t t

Put f(t) = logt – 1

t

∴∴∴∴ f′′′′(t) = 2

1 1+t t

∴∴∴∴ I = ( ) ( )′ ∫ t +e f t f t dt = etf(t) + c

= ( )

logx 1log log – +

loge x c

x

∴∴∴∴ I = x log (log x) – +log

xc

x

Q-3) tan∫∫∫∫ xdx

Ans. Let I = tanx dx∫Put tan x = t2

∴∴∴∴ sec2 x dx = 2t dt

dx = 2

2

1+ tan

t dt

x

= 4

2

1+

tdt

t

∴∴∴∴ I =2

4

2

1+

tt dt

t⋅∫

I =

2

4

2

1+

tdt

t∫

=( ) ( )2 2

4

+1 + –1

1+

t tdt

t∫

=

2

4

+1

+1

t

t∫ dt +

2

4

–1

+1

t

t∫ dt

= I1 + I

2...(i)

Here I1

=

2

4

+1

+1

t

t∫ dt

Divide numerator and denominator by t2

=2

2

2

11+

1+

t

tt

∫ dt

=2

2

11+

1– + 2

t

tt

∫ dt

Put1

– =t zt

∴∴∴∴ 2

11+

t

dt = dz

I1=

( )2

2

1

+ 2dz

z∫

=–11

tan +2 2

zc

=–1

1

1–

1tan +

2 2

tt c

=

2–1

1

1 –1tan +

2 2

tc

t

⋅

=–1

1

1 tan –1tan +

2 2tan

xc

x

...(ii)

Let I2=

2

4

–1

+1

t

t∫ dt

Divide numerator and denominator by t2

∴∴∴∴ I2=

2

2

2

11 –

1+

t

tt

∫ dt

=

2

2

11 –

1+ – 2

t

tt

∫ dt



Put1

+ =tt

θ ∴∴∴∴ 2

11–

t

dt = dθ

∴∴∴∴ I2=

( )2

2

1

– 2

dθ

θ

⋅∫

= 2

1 – 2log +

2 2 + 2c

θ

θ

= 2

1+ – 2

1log +

12 2 + + 2

tt

c

tt

=2

22

1 – 2 +1log +

2 2 + 2 +1

t tc

t t

= 2

1 tan – 2tan +1log +

2 2 tan + 2tan +1

x xc

x x

From (ii) and (iii), equation (i) becomes

I =–11 tan –1

tan +2 2tan

x

x

1 tan – 2tan +1log +

2 2 tan + 2tan +1

x xc

x x

Q-4)1

sin + tandx

x x∫∫∫∫

Ans. I =1

sin + tandx

x x∫

=cos

sin cos +sin

xdx

x x x∫

=( )

cos

sin 1+ cos

xdx

x x∫

=cos

cosec1+ cos

xx dx

x∫

=( )2cos /2 –1

cosec2cos /2

xx dx

x∫2

2

=1

cosec 1 –2cos /2

x dxx

∫ 2

=cosec

cosec –2 cos / 2

xx dx

x∫ 2

=cosec

cosec –1+ cos

xx dx dx

x∫ ∫

= log |cosec x – cot x|

( )

1–sin 1+cos

dxx x∫

Let I1

= ( )

1

sin 1+ cosdx

x x∫

= ( )

sin

sin 1+ cos

xdx

x x∫ 2

= ( ) ( )

sin

1 – cos 1+ cos

xdx

x x∫ 2

Put cos x = t ∴∴∴∴ – sin x dx = dt

I1

=( ) ( )

–

1 – 1+

dt

t t∫ 2

= ( ) ( ) ( )

–1

1 – 1+ 1+dt

t t t∫

=( ) ( )

–1

1 – 1+dx

t t∫ 2

( ) ( )

–1

1 – 1+t t2

=( ) ( ) ( )

+ +1+ 1+ 1 –

A B C

t t t

–1 = A (1 + t) (1 – t) + B (1 – t)

+ C (1 + t)2 (2)

Put t = 1 in (2)

–1 = C(2)2 ⇒⇒⇒⇒ C = –1

4

Put t = –1 in (2)

–1 = B (1 + 1) B = –1

2

Compaing co-efficents of t2

A + C = 0 ∴∴∴∴ A = 1

4

I1

=( ) ( )

1/4 –1/2 –1/4+ +

1+ 1 –1+dt

t tt

∫ 2

=

( )

1 1 –1 1 1 1–

4 1+ t 2 4 1 –1+dt dt dt

tt∫ ∫ ∫2

= ( )

1 1 1log 1+ + + log 1 – +

4 2 1+ 4t t c

t



I1

=( )

1 1log 1+ cos +

4 2 1+ cosx

x

1+ cos 1 – cos +4

x c

I =

1log cosex – cot – log 1+ cos

4x x x

= ( )

–1 1– log 1 – cos +

2 1+ cos 4x c

x

Q-5) (((( ))))2

log 2 log ∫∫∫∫

xe

x x x dxx

+

Ans. Let I = ( )2

x

log + 2loge

x x x dxx ∫

= ( )2x 2log

log +x

e x dxx

∫

Put f (x) = (logx)2

∴∴∴∴ f ′′′′ (x) =

d

dx(logx)2 = ( ) ( )2 log log

dx x

dx⋅

=2log x

x

∴∴∴∴ I = ( ) ( )x +e f x f x dx′ ∫= ex .f (x) + c = ex . (logx)2 + c .

AA–8

Q-1) (((( ))))3

1

sin sin +dx

x x αααα⋅⋅⋅⋅∫∫∫∫

Ans. Let I =( )α⋅∫ 3sin sin +

dxdx

x x

=( )α α⋅∫ 3sin sin cos + cos sin

dx

x x x

=( )α α∫ 4sin cos + cot sin

dx

x x

=α α⋅∫

2cosec d

cos +cot sin

x x

x

Put cos α + cot x.sin α = t

∴∴∴∴ –cosec2x.sin α dx = dt

∴∴∴∴ cosec2x dx = –αsin

dt

∴∴∴∴ I = α

∫

1 –

sin

dt

t=

α∫ ∫ –1 2–1

sint dt

=α

⋅1 2–1

+sin 1 2

tc

Q-2)

3

11

cos

sin

xdx

x∫∫∫∫

Ans. I = ∫3

11

cos

sin

xdx

x

= ⋅∫3

3 8

cos 1

sin sin

xdx

x x

= ⋅∫ 3

4

1cot

sinx dx

x

= ⋅ ⋅∫3

22 2

1 1cot

sin sinx dx

x x

= ⋅ ⋅∫3

2 22cot cosec cosecx x xdx

Put cotx = t

∴∴∴∴ –cosec2x dx = dt

∴∴∴∴ cosec2x dx = –dt

∴∴∴∴ I = ( ) ( )∫3

22t 1+ t –dt = – ∫

3 7

2 2t + t dt

= – ∫ ∫3 7

2 2t – tdt dt

=

3 7+1 +1

2 2– t t– +

3 7+1 +1

2 2

c

=

5 9

2 22 2

– t – t +5 9

c

∴∴∴∴ I =

5 9

2 22 2

– cot – cot +5 9

x x c



Q-3) 2

1

sin cos + 2cosdx

x x x⋅⋅⋅⋅∫∫∫∫

Ans. I =⋅∫ 2

1

sin cos +2cosdx

x x x

Divide by cos2x

I = ∫2sec

tan +2

xdx

x

f(x) = tanx + 2 f′′′′(x) = sec2x

I = log tan +2 +x c

Q-4)

2

– –

x

x x

edx

e e∫∫∫∫

Ans. I = ∫2

–

x

x x–

edx

e e

= ∫2x

x

x

1–

edx

ee

=

( )∫2

2

x

x

x

1 –

edx

e

e

=

( )∫2

2

x x

x1 –

e edx

e

=

( )

⋅

∫2 2

2

x x

x1 –

e edx

e

= ( )∫ 2

x

x1 –

edx

e

Put ex = t ∴∴∴∴ ex dx = dt

I = ∫ 2

1

1 –dx

t

= sin–1 t + c

= ∫2

–

x

x x–

edx

e e

= sin–1 (ex ) + c

Q-5) 1 secx dx∫∫∫∫ +

Ans. I = ∫ 1+sec x dx

= ∫1

1+sec

dxx

= ∫cos +1

cos

xdx

x= ∫

22cos2

cos

x

dxx

= ∫ 2

2 cos2

1 – 2sin2

x

x

Put sin2

x = t

∴∴∴∴

1cos

2 2

x dx = dt

∴∴∴∴ cos2

xdx = 2dt

∴∴∴∴ I = ∫ 2

12

1 – 2t × × × × 2dt

=

∫2

12 2

12 –

2

dt

t

=

∫ 2

2

2 2 1

2 1–

2

dt

t

= –12sin +

1

2

tc

∴∴∴∴ I =

–12sin 2sin +2

xc

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Transcript of sin .cos ∫ - MT Educare,mteducaresouth.com/images/maths/homework/Indefinite-Integration-13... ·...