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Transcript of Integration e.g.
8/3/2019 Integration e.g.
http://slidepdf.com/reader/full/integration-eg 1/11
Examples In Integration
2 2I x a x dx= −∫ let
2 2 du duu a x 2x dx
dx 2x= − ⇒ = − ⇒ =
−, then
duI x u
2x= ⋅
−∫
3
1 3 322 22 2 2
1 1 u 1 1u du c u c (a x ) c
32 2 3 32
= − = − + = − + = − − +
∫
-----------------------------------------------------------------------------------
-------------------------------------------
2
2 3
xI dx
a x=
+∫ let
2 3 2
2
du duu a x 3x dx
dx 3x= + ⇒ = ⇒ = , then
2
2
x duI
3xu= ⋅∫
11 1 12
2 32 2 21 1 u 2 2
u du c u c (a x ) c13 3 3 32
− = = + = + = + +
∫
-----------------------------------------------------------------------------------
-------------------------------------------
2I sin xcosxdx= ∫ let2 du
u sin x 2cosxsinxdx
= ⇒ = &dv
cosx v sinxdx
= ⇒ = , then
33 2 3 3 sin x
I sin x 2 sin xcosxdx sin x 2I 3I sin x I c3
= − = − ⇒ = ⇒ = +∫ -----------------------------------------------------------------------------------
-------------------------------------------
3xI dxx 1= +∫ , by algebraic long division
3
2
x 1x x 1x 1 x 1= − + −+ + , then this
implies that2 3
2 1 x xI 1dx xdx xdx dx x ln(x 1) c
x 1 2 3= − + − = − + − + +
+∫ ∫ ∫ ∫ -----------------------------------------------------------------------------------
-------------------------------------------
2 2sec 1 3sec 1I d d ln(1 3tan ) c
1 3tan 3 1 3tan 3
θ θ= θ = θ = + θ +
+ θ + θ∫ ∫ , sincef'(x)
dx ln(f(x)) cf(x)
= +∫ -----------------------------------------------------------------------------------
-------------------------------------------
2
secx(secx tanx) sec x secxtanxI secxdx dx dx ln(secx tanx) csecx tanx secx tanx
+ += = = = + ++ +∫ ∫ ∫
-----------------------------------------------------------------------------------
-------------------------------------------
1I dx
1 cosx=
+∫ let 2
x 2t tan dx dt
2 1 t
= ⇒ = + , then
2
2
2
2
1 tI dt1 t
11 t
+=−
++
∫
2
2 2
2 2 x 1 xdt dt 1dt t c tan c tanx 1 tan c
1 t 1 t 2 2 2 2
= = = = + = + = − + + + − ∫ ∫ ∫ , from
8/3/2019 Integration e.g.
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2
2tanAtan2A
1 tan A=
−,
2
2
xsin
1 1 1 cosx2I tan 1 c tanx 1 c
x2 2 1 cosxcos
2
− = − + = − + +
, from
2 2
cos2A 2cos A 1 1 2sin A= − = − , so now we have1 1 cosx 1 cosx
I tanx c2 1 cosx 1 cosx
+ − = − + + + 1 2cosx cosx sinx cosx sinxtanx c tanx c c c
2 1 cosx 1 cosx cosx 1 cosx 1 cosx
= + = + = + = + + + + +
2 2
sinx(1 cosx) sinx sinxcosx sinx(1 cosx) 1 cosxc c c c
(1 cosx)(1 cosx) 1 cos x sin x sinx
− − − −= + = + = + = +
+ − −,
therefore we finally have,1 cosx
I c cotx cosecx csinx sinx
= − + = − + +
-----------------------------------------------------------------------------------
-------------------------------------------
2 2
1I dx
x a=
+∫ let 2x atan dx asec d= θ ⇒ = θ θ , then2
2 2 2
asecI d
a tan a
θ= θ
θ +∫
2 21
2 2 2
asec 1 sec 1 1 1 xd d 1d c tan c
a (tan 1) a sec a a a a
−θ θ = θ = θ = θ = θ + = + θ + θ ∫ ∫ ∫ -----------------------------------------------------------------------------------
-------------------------------------------
2 2
1I dx
a x=
−
∫ let x asin dx acos d= θ ⇒ = θ θ , then2 2 2
acosI d
a a sin
θ= θ
− θ
∫
1
2 2
acos cos cos xd d d 1d c sin c
cos aa 1 sin cos
−θ θ θ = θ = θ = θ = θ = θ + = + θ − θ θ∫ ∫ ∫ ∫
-----------------------------------------------------------------------------------
-------------------------------------------
2 2
1I dx
x a=
−∫ let x asec dx asec tan d= θ ⇒ = θ θ θ , then 2 2 2
asec tanI d
a sec a
θ θ= θ
θ −∫
2 2 2
asec tan 1 sec tan 1 sec 1 1d d d d
a (sec 1) a tan a tan a sin
θ θ θ θ θ= θ = θ = θ = θ
θ − θ θ θ∫ ∫ ∫ ∫ let t tan2
θ =
22d dt
1 t⇒ θ =
+, so 2
2
1 1 2 1 1 1 1I dt dt lnt c ln tan c2ta 1 t a t a a 2
1 t
θ = ⋅ = = + = + +
+∫ ∫
2
2
2
sin1 1 1 1 2ln tan 1 tan c ln tan 1 c
a 2 2 a 2cos
2
θ θ = θ − + = θ − + θ
1 1 1 cos 1 1 1 cos 1 cosln tan 1 c tan c
a 2 1 cos a 2 1 cos 1 cos
− θ + θ − θ = θ − + = θ − + + θ + θ + θ
1 1 2cos 1 cos 1 sin cosln tan c ln tan c ln ca 2 1 cos a 1 cos a cos 1 cos
θ θ θ θ = θ + = θ + = ⋅ + + θ + θ θ + θ
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21 sin 1 tan cos 1 sec 1 cosln c ln c ln c
a 1 cos a 1 cos a 1 cos
θ θ θ θ − ⋅ θ = + = + = + + θ + θ + θ
( ) ( ){ }2
21 sec 1 1ln c ln sec 1 ln sec 1 c
a sec 1 a
θ −= + = θ − − θ + +
θ +
( ) ( ) ( ) ( )( ) ( )21 1 1 1ln sec 1 ln sec 1 c ln sec 1 sec 1 ln sec 1 c
a 2 a 2
= θ − − θ + + = θ − θ + − θ + +
( ) ( )( ) ( ){ }1 x x x 1
ln 1 1 2ln 1 c ln x a x a 2ln x a c2a a a a 2a
= − + − + + = − + − + +
( ) ( )
( )2
x a x a1ln c
2a x a
− + = + +
, we finally have 1 x a
I ln c2a x a
− = + +
-----------------------------------------------------------------------------------
-------------------------------------------
2 2
1I dx
x a=
+∫ let 2x atan dx asec d= θ ⇒ = θ θ , then
2
2 2 2
asecI d
a tan a
θ= θ
θ +∫
( )2 2
2
asec secd d sec d ln tan sec c
seca tan 1
θ θ= θ = θ = θ θ = θ + θ +
θθ +∫ ∫ ∫
( ) ( )2
2 2 2
2
x xln tan tan 1 c ln 1 c ln x x a c
a a
= θ + θ + + = + + + = + + +
-----------------------------------------------------------------------------------
-------------------------------------------
2 2
1I dx
x a=
−∫ let x asec dx asec tan d= θ ⇒ = θ θ θ , then
2 2 2
asec tanI d
a sec a
θ θ= θ
θ −∫
( )2 2
asec tan sec tand d sec d ln sec tan c
a sec 1 tan
θ θ θ θ= θ = θ = θ θ = θ + θ +
θ − θ∫ ∫ ∫
( ) ( )2
2 2 2
2
x xln sec sec 1 c ln 1 c ln x x a c
a a
= θ + θ − + = + − + = + − +
-----------------------------------------------------------------------------------
-------------------------------------------
2
1I dx
x 2x 5=
+ +∫ completing the square,
1
2
1 1 x 1I dx tan c
(x 1) 4 2 2
− + = = + + + ∫ -----------------------------------------------------------------------------------
-------------------------------------------
8/3/2019 Integration e.g.
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2
1I dx
2 x x=
+ −∫ manipulate in order to complete the square,
2
1I dx
x x 2=
− − − ∫
1 1
22
1x
1 1 2x 12dx dx sin c sin c3 39 11 9
xx 24 22 4
− −
− − = = = + = + − −− − −
∫ ∫
2
sinI d
9 4cos
θ= θ
− θ∫ let
12cos x d
2sinθ = ⇒ θ = −
θ, then
2
1 sin dxI
2 sin9 x
θ= − ⋅
θ−∫
1 1
2
1 dx 1 x 1 2cossin c sin c
2 2 3 2 39 x
− − θ = − = − + = − + −
∫ -----------------------------------------------------------------------------------
-------------------------------------------
( )2
2 5 2 2 2 2 2I sin xcos xdx sin xcos xcos xcosxdx sin x 1 sin x cosxdx= = = −∫ ∫ ∫
( ) ( )2 2 4 2 4 6sin x 1 2sin x sin x cosxdx sin x 2sin x sin x cosxdx= − + = − +∫ ∫ let u sinx=
du ducosx dx
dx cosx⇒ = ⇒ = , then ( ) ( )2 4 6 2 4 6du
I u 2u u cosx u 2u u ducosx
= − + ⋅ = − +∫ ∫ ,
so by normal power integration,3 5 7 3 5 7u 2u u sin x 2sin x sin x
I c c
3 5 7 3 5 7
= − + + = − + +
-----------------------------------------------------------------------------------
-------------------------------------------
( )3 2 2I cos xdx cos xcosxdx 1 sin x cosxdx= = = −∫ ∫ ∫ , letdu
u sinx dxcosx
= ⇒ = , then
( ) ( )3 3
2 2du u sin xI 1 u cosx 1 u du u c sinx c
cosx 3 3= − ⋅ = − = − + = − +∫ ∫
-----------------------------------------------------------------------------------
-------------------------------------------
( ) ( )2 2 2 2 2 4I sin xcos xdx sin x 1 sin x dx sin x sin x dx= = − = −∫ ∫ ∫ , to integrate 4sin x we
need a formula with this included, so consider( )cos4x cos 2x 2x cos2xcos2x sin2xsin2x= + = −
( ) ( ) ( ) ( )2 2 2 4 2 21 2sin x 1 2sin x 2sinxcosx 2sinxcosx 1 4sin x 4sin x 4sin xcos x= − − − = − + −
( )2 4 2 2 2 4 2 41 4sin x 4sin x 4sin x 1 sin x 1 4sin x 4sin x 4sin x 4sin x= − + − − = − + − +
2 4 4 2 4 21 11 8sin x 8sin x 8sin x cos4x 1 8sin x sin x cos4x sin x
8 8= − + ⇒ = − + ⇒ = − + ,
( )2 21 1 1 x sin4xI sin x cos4x sin x dx 1 cos4x dx c
8 8 8 8 32
= − − + = − = − +
∫ ∫ -----------------------------------------------------------------------------------
-------------------------------------------
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2I cos xdx= ∫ , ( 2cos2x 2cos x 1= − ) therefore, ( )
1 x 1I cos2x 1 dx sin2x c
2 2 4= + = + +∫
-----------------------------------------------------------------------------------
-------------------------------------------
2 2
I a x dx= −∫
letx asin dx acos d= θ ⇒ = θ θ
, then
2 2 2
I a a sin acos d= − θ ⋅ θ θ∫
( )2 2
2 2 2 2 a sin2 aa 1 sin cos d a cos d c sin cos c
2 2 2
θ = − θ θ θ = θ θ = + θ + = θ θ + θ + ∫ ∫
2 2 22 1 1
2
a x x a x x x1 sin sin c 1 sin c
2 a a 2 a a a
− − = ⋅ − θ + + = ⋅ − + +
2 22 2 1 2 2 1a x 1 x x a x
a x sin c a x sin c2 a a a 2 2 a
− − = ⋅ ⋅ − + + = − + +
-----------------------------------------------------------------------------------
-------------------------------------------
( )3
2 2 2
1I dx
a x
=−
∫ let x asin dx acos d= θ ⇒ = θ θ , then( )
32 2 2 2
acosI d
a a sin
θ= θ− θ
( ) ( )
2
3 32 2 2 2 2 22 22 2
1 cos 1 cos 1 1 1 tand d d sec c
a a a cos a a1 sin cos
θ θ θ= θ = θ = θ = θ = +
θ− θ θ
∫ ∫ ∫ ∫
2 3 2 2 2 2 23
2
sin x x xc c c c
a cos a 1 sin x a a xa 1
a
θ= + = + = + = +
θ − θ −−
-----------------------------------------------------------------------------------
-------------------------------------------
2
1I dx
x x 1=
+∫ let 2x tan dx sec d= θ ⇒ = θ θ , then
2
2
secI d
tan tan 1
θ= θ
θ θ +∫
2sec sec 1d d d
tan sec tan sin
θ θ= θ = θ = θ
θ θ θ θ∫ ∫ ∫ let 2
2t tan d dt
2 1 t
θ = ⇒ θ = + , then
2
2
1 2 2 1I dt dt dt lnt c ln tan c
2t 1 t 2t t 2
1 t
θ = ⋅ = = = + = + + +
∫ ∫ ∫ , now from double
angle formulae
2
2
2
sin1 1 2
I ln tan 1 tan c ln tan 1 c2 2 2
cos2
θ θ = θ − + = θ − + θ
1 1 cos 1 2cos cosln tan 1 c ln tan c ln tan c
2 1 cos 2 1 cos 1 cos
− θ θ θ = θ − + = θ + = θ + + θ + θ + θ
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( )
( )
2
22 2 2
x x 1 1tan tan xln c ln c ln c ln c
sec 1 tan 1 1 x 1 1 x 1 1
+ − θ θ = + = + = + = + θ + θ + + + + + −
( )2
2
2
x x 1 1 x 1 1ln c ln cx x
+ − + − = + = + -----------------------------------------------------------------------------------
-------------------------------------------
3 2
2x 3I dx
x x 2x
+=
+ −∫ now if 3 2f(x) x x 2x= + − , then f(1) 0 (x 1)= ⇒ − is a
factor, so by long division 3 2 2x x 2x (x )(x 2x) (x 1)x(x 2)+ − = − + = − + , by
partial fractions we have that5 1 3 1 1 1 3 5 1
I dx dx dx lnx ln(x 1) ln(x 2) c3 x 1 2 x 6 x 2 2 3 6= − + = − + − + + +− +∫ ∫ ∫ -----------------------------------------------------------------------------------
-------------------------------------------
2
3
x 1I dx
(x 1)
+=
−∫ by partial fractions 3 2
1 1 1I 2 dx 2 dx dx
(x 1) (x 1) x 1= + +
− − −∫ ∫ ∫ ,
then2 1
2
2 2 1 2I (x 1) (x 1) ln(x 1) c ln(x 1) c
2 1 (x 1) x 1
− −= − − − − + − + = − − + − +− −
-----------------------------------------------------------------------------------
-------------------------------------------
2xI dx(x 1)(x 4)
= + +∫ by partial fractions
2 2
1 1 1 x 4 1I dx dx dx
5 x 1 5 x 4 5 x 4= − + +
+ + +∫ ∫ ∫
2 1
2 2
1 1 1 2x 4 1 1 1 2 xdx dx dx ln(x 1) ln(x 4) tan c
5 x 1 10 x 4 5 x 4 5 10 5 2
− = − + + = − + + + + + + + + ∫ ∫ ∫
{ }2
2 1 1
2
1 2 x 1 x 4 2 xln(x 4) 2ln(x 1) tan c ln tan c
10 5 2 10 (x 1) 5 2
− − + = + − + + + = + + + -----------------------------------------------------------------------------------
-------------------------------------------
( )
3
22
2x x 3I dx
x 1
+ +=
+∫ by partial fractions
( )22
2
2x 3 xI dx
x 1 x 1
− = + + +
∫
( ) ( )2 22
2 2
2x 1 xdx 3 dx dx
x 1 x 1 x 1= + −
+ + +∫ ∫ ∫ , now take each integrand
separately so ( )2
2
2xdx ln x 1
x 1= +
+∫ ,( )
22
xdx
x 1+∫ let
2 duu x 1 dx
2x= + ⇒ = ,
then2
1 1du2 u∫
( )
1
21 u 12 1 2 x 1
−
= = − − + ,
( )2
2
13 dx
x 1+∫ let
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2x tan dx sec d= θ ⇒ = θ θ , then
( )
2 22
2 4 22
sec sec 1 33 d 3 d 3 d 3 cos d (cos2 1)d
sec sec 2tan 1
θ θθ = θ = θ = θ θ = θ + θ
θ θθ +∫ ∫ ∫ ∫ ∫
2
2
3 sin2 3 3 3 tan
(sin cos ) (tan cos )2 2 2 2 2 sec
θ θ = + θ = θ θ + θ = θ θ + θ = + θ θ
1
2
3 xtan x
2 x 1
− = + + , ( )
( )2 1
2 2
3 x 1I ln x 1 tan x c
2 x 1 2 x 1
− = + + + + + + +
( )( ) ( )
( )( )
2 1 2 1
2 2 2
3x 1 3 1 3x 3ln x 1 tan x c ln x 1 tan x c
2 22 x 1 2 x 1 2 x 1
− −+= + + + + + = + + + +
+ + +
-----------------------------------------------------------------------------------
-------------------------------------------
1 sinxI dx
sinx(1 cosx)
+=
+∫ let
1
2
x 2t tan x 2tan t dx dt
2 1 t
− = ⇒ = ⇒ = +
, then
22 2
22 2 22
22 2
2t 2t1 1
2 1 t 2t1 t 1 tI dt 2tdt 2dt2t(1 t )1 t 2t(1 t ) 2t(1 t )2t 1 t
2t11 t1 t 1 t
+ + + ++ += ⋅ = ⋅ = ⋅−+ + + − − ++ ++ +
∫ ∫ ∫
( )
2 2 2 2
2 2
t 2t 1 t 2t 1 t 2t 1 t 2t 12dt 2dt dt dt dt dt
4t 2t 2t 2t 2t2t 1 t 1 t
+ + + + + += ⋅ = ⋅ = = + +
+ + −∫ ∫ ∫ ∫ ∫ ∫ 2
21 1 1 1 t 1 1 x x 1 xtdt 1dt dt t lnt c tan tan ln tan c
2 2 t 2 2 2 4 2 2 2 2
= + + = ⋅ + + + = + + +
∫ ∫ ∫ -----------------------------------------------------------------------------------
-------------------------------------------
1I dx
5 3cosx=
−∫ let 2
x 2t tan dx dt
2 1 t
= ⇒ = + , then 2 2
2
1 2I dt
3(1 t ) 1 t5
1 t
= ⋅− +−+
∫
2 2 2 2 2 22
2 2 2 1 1 1dt dt dt dt dt
15(1 t ) 3(1 t ) 5 5t 3 3t 8t 2 4t 1 4 t4
= = = = =+ − − + − + + + +
∫ ∫ ∫ ∫ ∫
1 1 11 1 t 1 1 xtan c tan (2t) c tan 2tan c
1 14 2 2 22 2
− − −
= ⋅ + = + = + -----------------------------------------------------------------------------------
-------------------------------------------
( )3 3
1I dx
x a x=
+∫ let3
2
dtt x dx
3x= ⇒ = , then ( ) ( )3 3 3
1 1 1 1I dt dt
3 3x a t t a t= =
+ +∫ ∫
, by partial fractions3
3 3 3 3 3 3
1 1 1 1 1 xI dt dt ln c
3a t 3a a t 3a a x
= − = + + +
∫ ∫ -----------------------------------------------------------------------------------
-------------------------------------------
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2I x lnxdx= ∫ letdu 1
u lnxdx x
= ⇒ = &3
2dv xx v
dx 3= ⇒ = , then by the parts
formula we have3 3 3 3
2x lnx 1 x lnx 1 x x 1I xdx c lnx c
3 3 3 3 3 3 3
= − = − ⋅ + = − + ∫
-----------------------------------------------------------------------------------
-------------------------------------------
I xarctanxdx= ∫ let 2
du 1u arctanx
dx 1 x= ⇒ =
+&
2dv xx v
dx 2= ⇒ = , then by the
parts formula2 2 2 2
2 2 2
x arctanx 1 x x arctanx 1 1 x 1I dx dx dx
2 2 1 x 2 2 1 x 1 x
+= − = − − + + +
∫ ∫ ∫
( )2 2
2
x arctanx 1 1 x arctanx 11dx dx x arctanx
2 2 1 x 2 2
= − − = − − + ∫ ∫
( )22 2 x 1x arctanx x 1 x 1 x x
arctanx arctanx arctanx c2 2 2 2 2 2 2 2
+ = − + = + − = − + -----------------------------------------------------------------------------------
-------------------------------------------
1I xsinxcosxdx xsin2xdx
2= =∫ ∫ let
duu x 1
dx= ⇒ = &
dv cos2xsin2x v
dx 2= = = − ,
then1 x 1 1 x
I cos2x cos2xdx sin2x cos2x c2 2 2 8 4
= − + = − + ∫
-----------------------------------------------------------------------------------
-------------------------------------------
2I ln xdx lnx lnxdx= = ⋅∫ ∫ let du 1u lnxdx x
= ⇒ = &
dvlnx v xlnx x x(lnx 1)
dx= ⇒ = − = − , then
( )2 2I lnx(xlnx x) (lnx 1)dx xln x 2xlnx 2x c x ln x 2lnx 2 c= − − − = − + + = − + +∫ -----------------------------------------------------------------------------------
-------------------------------------------
xI xa dx= ∫ letdu
u x 1dx
= ⇒ = &x
xdv aa v
dx lna= ⇒ = , then
x xxa a
I dxlna lna
= − ∫
x x x
xa 1 a a 1c x clna lna lna lna lna
= − ⋅ + = − + -----------------------------------------------------------------------------------
-------------------------------------------
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116 4
1
0 2
xI dx
1 x
=+
∫ let 4 3x t dx 4t dt= ⇒ = , x 0 t 0= ⇒ = , x 16 t 2= ⇒ = , then
2 2 43
2 2
0 0
t tI 4t dt 4 dt
1 t 1 t
= ⋅ =+ +∫ ∫
by long division4
2
2 2
t 1t 1
1 t 1 t= + −
+ +, then we
have
22 32 1
2
0 0
1 4tI 4 t 1dt 4tan t 4t 7.1
1 t 3
− = + − = + − ≅ + ∫
-----------------------------------------------------------------------------------
-------------------------------------------
2
1I dx
1 x x=
+ +∫ let ( ) ( ) ( )
1 12 22 2
dt 1t x 1 x x 1 2x 1 1 x x
dx 2
−− = + + ⇒ − = + + + ⇒
2
2 2
dt 2x 1 2x 1 2 1 x x 2x 1 2(t x) 2(t x)1 dx dt
dx 2(t x) 2x 1 2(t x)2 1 x x 2 1 x x
+ + + + + + + − −= + = = ⇒ =
− + + −+ + + +
,
then ( ) ( )22 2I dt dt ln 2t 1 c ln 2x 1 2 1 x x c
2x 1 2t 2x 2t 1= = = + + = + + + + +
+ + − +∫ ∫ -----------------------------------------------------------------------------------
-------------------------------------------
2 2
4
a xI dx
x
−= ∫ let
2 2
2 2
1 1 dt 1t x dx xdt
x t dx x= ⇒ = ⇒ = − ⇒ = − , then it follows
that
2
22 2
2
1a
tI dt t a t 1 dt1
t
−= − = − − ⋅∫ ∫ let
2 2
2
duu a t 1 dt
2a t= − ⇒ = , therefore
( )
3 33 22 2
2 2 22 2 2 2 2
1 1 u 1 1 aI u du c a t 1 c 1 c
32a 2a 3a 3a x
2
= − ⋅ = − + = − − + = − − +
∫ , now we
finally have the following result; ( )3
2 2 2
2 2
a xI c
3a x
−= − +
-----------------------------------------------------------------------------------
-------------------------------------------
2x x
1I dxe 2e
=−∫ let
x
x
dtt e :lnt x dxe
= = ⇒ = , then 2 3 2
1 dt 1I dtt 2t t t 2t
= ⋅ =− −∫ ∫
2
1dt
t (t 2)=
−∫ now by the partial fraction theorem we have,
2
1 1 1 t 2I dt dt
4 t 2 4 t
+= −
−∫ ∫
( ) ( )x
2 x
1 1 1 1 1 1 1 1 1 1 x 1dt dt dt ln t 2 lnt c ln e 2 c
4 t 2 4 t 2 t 4 4 2t 2e 4 4= − − = − − + + = − + − +
−∫ ∫ ∫ -----------------------------------------------------------------------------------
-------------------------------------------
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( )
2x
1x 4
eI dx
e 1
=+
∫ letx
x
dtt e 1 dx
e= + ⇒ = , then
2x x
1 1 1x
4 4 4
e dt e t 1I dt dt
et t t
−= ⋅ = =∫ ∫ ∫
( ) ( ) ( )3 1 7 3 3 3
x x4 4 4 4 4 44 4 4 4
t dt t dt t t c 3t 7 t c 3e 4 e 1 c
7 3 21 21
− = − = − + = − + = − + +
∫ ∫ -----------------------------------------------------------------------------------
-------------------------------------------
ln5 x x
x
0
e e 1I dx
e 3
−=
+∫ 2 x
x
2tt e 1 dx dt
e= − ⇒ = , x 0 t 0= ⇒ = , x ln5 t 2= ⇒ = , then
2 2
2
0
tI 2 dt
t 4=
+∫ let 2t 2tan dt 2sec d= θ ⇒ = θ θ , then
2 2 2
2
0
4tan 2secI 2 d
4tan 4
θ⋅ θ= θ
θ +∫
[ ]22 22 2
22 1
2 0
0 0 0
8tan sec t t2 d 4 tan d 4 tan 4 tan 4
4sec 2 2
− θ ⋅ θ = θ = θ θ = θ − θ = − = − π
θ
∫ ∫ -----------------------------------------------------------------------------------
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ln2x
0
I e 1 dx= − ⋅∫ letx
x
dtt e 1 dx
e= − ⇒ = , x 0 t 0= ⇒ = , x ln2 t 1= ⇒ = , then
1 1
x
0 0
t tI dt dt
e t 1= =
+∫ ∫ let 2 2t tan dt 2sec tan d= θ ⇒ = θ θ θ , then
1 2 2
2
0
tan secI 2 d
tan 1
θ⋅ θ= θ
θ +∫
[ ] ( )1 1
12 1
00
0
42 tan d 2 tan 2 t tan t 2 1 2
4 2 2
− π π − π = θ θ = θ − θ = − = − = − = ∫ -----------------------------------------------------------------------------------
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1 lnxI dx
x
+= ∫ let t 1 lnx dx xdt= + ⇒ = , then
( )3 3
2 22 2
I t dt t c 1 lnx c3 3
= ⋅ = + = + +∫ -----------------------------------------------------------------------------------
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( )4
0
sin cosI d
3 sin2
π
θ + θ= θ
+ θ∫ letdt
t sin cos dsin cos
= θ − θ ⇒ θ =θ + θ
, t 04
πθ = ⇒ = ,
0 t 1θ = ⇒ = − , then0
1
dtI
3 sin2−
=+ θ∫ , now consider the following
( )22t sin cos= θ − θ
8/3/2019 Integration e.g.
http://slidepdf.com/reader/full/integration-eg 11/11
( )
0 02 2 2
221 1
dt dtsin cos 2sin cos 1 t sin2 I
4 t3 1 t− −
= θ + θ − θ θ ⇒ − = θ ⇒ = =−+ −∫ ∫ , so
therefore we have that
0 1
1
1 2 t 1 1 1 1 ln3I ln ln1 ln ln
4 2 t 4 3 4 3 4
−
−
+ = = − = =
− -----------------------------------------------------------------------------------
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( ) ( ){ } ( )1sinx sinx
I tanxdx dx dx ln cosx c ln cosx c ln secx ccosx cosx
−−= = = − = − + = + = +∫ ∫ ∫
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These are a good selection of integrals, the solutions cover the main techniques that are required to attack most integration problems.