Mathematics. Session Indefinite Integrals - 3 Session Objectives Three Standard Integrals ...
-
Upload
aileen-butler -
Category
Documents
-
view
232 -
download
4
Transcript of Mathematics. Session Indefinite Integrals - 3 Session Objectives Three Standard Integrals ...
Mathematics
Session
Indefinite Integrals - 3
Session Objectives
Three Standard Integrals
2 2 2 2 2 2a - x dx, x +a dx, x - a dx
Integrals of the form 2ax +bx+c dx
2px+q ax +bx+c dx Integrals of the form
1 1dx, dx
a+bsinx a+bcosx Integrals of the form
Integration Through Partial Fractions
Class Exercise
Three Standard Integrals
2
2 2 2 2 -1x a x1 a - x dx = a - x + sin +C
2 2 aPut x =asin or acos
2
2 2 2 2 2 2e
x a2 x +a dx = x +a + log x+ x +a +C
2 2Put x =atan or acot
2
2 2 2 2 2 2e
x a3 x - a dx = x - a - log x+ x - a +C
2 2
Put x =asec or acosec
Integrals of the Form
2ax +bx+c dx
Reduce the given integral to one of the following forms:
2 2 2 2 2 2a - x dx or x +a dx or x - a dx
Example-1
2Evaluate: 16x +25 dx2Solution: Let I = 16x +25 dx
22 5
= 4 x + dx4
2
2 22 2
e
5x 5 54
= 4 x + + log x+ x + +C2 4 2 4
2 2e
25 25 25=2x x + + log x+ x + +C
16 8 16
Example - 2
2Evaluate: x +8x+4 dx2Solution: Let I = x +8x+4 dx
2= x +8x+16 - 16+4 dx
2= x+4 - 12 dx
22= x+4 - 2 3 dx
Solution Cont.
22 22 2
e
2 3x+4= x+4 - 2 3 - log x+4 + x+4 - 2 3 +C
2 2
22 2 2 2 2 2
ex a
x - a dx = x - a - log x+ x - a +C2 2
2
2e
x+4 x +8x+4= - 6log x+4 + x +8x+4 +C
2
Example - 3
2Evaluate : 7x - 10- x dx 2Solution: Let I = 7x - 10- x dx
2= -10- x - 7x dx 249 49= -10+ - x - 7x+ dx
4 4
2 23 7= - x - dx
2 2
2 -11 9 2x - 7= 2x - 7 7x - 10- x + sin +C
4 8 3
2
2 2-1
7 3 7x - x -3 72 2 2= - x - + sin +C32 2 2 22
Integrals of the Form
2px+q ax +bx+c dxWe use the following method:
2di Express px+q= A ax +bx+c +B
dx
px+q= A 2ax+b +B Identity
(ii) Obtain the values of A and B by comparing the coefficients of like powers of x. Then the integral reduces to
2 2A 2ax+b ax +bx+c dx+B ax +bx+c dx
2iii To evaluate first integral, put ax +bx+c= t
2ax+b dx =dt and second integral by the
method discussed earlier.
Example - 4
2Evaluate: x x+x dx
2Solution: Let I = x x+x dx
2dPutting x = A x + x +B
dx
x = A 1+2x +B Identity
1 1A = , B = -
2 2
2 21 1I = 1+2x x+x dx - x+x dx
2 2
Solution Cont.
2 21 2Let I = 1+2x x+x dx and I = x+x dx
21I = 1+2x x+x dx
2Putting x + x = t 1+2x dx = dt
3 3
22 21
2 2I = t dt= t = x+x
3 3
2 22
1 1I = x+x dx = x +x+ - dx
4 4
2 21 1= x+ - dx
2 2
Solution Cont.
2 2 2 2
e1 1 1 1 1 1 1 1 1
= x+ x+ - - × log x+ + x+ -2 2 2 2 2 4 2 2 2
2 2 2 2
e1 1 1 1 1 1 1 1
= x+ x+ - - log x+ + x+ -2 2 2 2 8 2 2 2
3 2 2 2 2
2 2e
1 2 1 1 1 1 1 1 1 1 1I = × x+x - x+ x+ - - log x+ + x+ - +C
2 3 2 2 2 2 2 8 2 2 2
3
2 2 22e
1 1 1 1 1= x+x - x+ x+x - log x+ + x+x +C
3 4 2 4 2
Integrals of the Form
1 1dx, dx
a+bsinx a+bcosx We use the following method:
2
2 2
x x2tan 1- tan
2 2i Write sinx = , cosx =x x
1+tan 1+tan2 2
2
2
x 1 xii Putting tan = t sec dx =dt, we get
2 2 21
the integral in the form dxat +bt+c
(iii) Now, we evaluate the integral by the method discussed earlier.
Example - 51
Evaluate: dx4 cosx - 1
1Solution: Let I = dx
4 cosx - 1
2
2
x1- tan
2Putting cosx = , we getx
1+tan2
2
2 22
2
xsec1 2I = dx = dx
x xx 4- 4tan - 1- tan1- tan2 224 - 1
x1+tan
2
Solution Cont.
2
2
xsec
2= dxx
3- 5tan2
2 2x 1 x xPutting tan = t sec dx = dt sec dx = 2dt
2 2 2 2
222
2 dt 2 dtI = =
35 5- t 3- t5
5
Solution Cont.
e
3+t
2 1 5= × log +C5 3 3
2 - t5 5
e
x3+ 5tan1 2= log +C
x15 3 - 5tan2
Integration Through Partial Fractions (Type – 1)
When denominator is non-repeated linear factors
f(x) A B C
Let = + +x- a x - b x - c x - a x - b x - c
where A, B, C are constants and can be calculated by equating the numerator on RHS to numerator on LHS and then substituting x = a, b, c, ... or by comparing the coefficients of like powers of x.
Example - 6
2x+1
Evaluate: dxx+1 x+2
2x+1
Solution: Let I = dxx+1 x+2
2x+1 A B
Let = +x+1 x+2 x+1 x+2
x+2 A+ x+1 B2x+1
=x+1 x+2 x+1 x+2
2x +1 = A x +2 +B x +1 Identity
Solution Cont.
Putting x = -1, - 2, we get
A = -1 and B = 3
dx dxI =- +3
x+1 x+2
e e= -log x +1 +3log x +2 +C
3
ex+2
=log +Cx+1
Type - 2
When denominator is repeated linear factors
2 3 2 2 3
f(x) A B C D E FLet = + + + + +
x- a x - b x - c(x - a) (x - b) (x - c) (x - b) (x - c) x - c
where A, B, C, D, E and F are constants and value of the constants are calculated by substitution as in method (1) and remaining are obtainedby comparing coefficients of equal powers of x on both sides.
Example - 7
23x+1
Evaluate: dxx - 2 x+2
23x+1
Solution: Let I = dxx - 2 x+2
2 2
3x+1 A B CLet = + +
x- 2 x+2x - 2 x+2 x - 2
23x+1= A x - 2 x+2 +B x+2 +C x - 2
223x+1= A x - 4 +B x+2 +C x - 2 Identity
7Putting x =2, we get 7= 4B B =
4
Solution Cont.-5
Putting x =-2, we get - 5=16C C =16
2Comparing coefficients of x on both sides, we get
5A+C =0 A =-C A =
16
2 2
3x+1 5 7 5= + -
16 x - 2 16 x+2x - 2 x+2 4 x - 2
2
5 1 7 1 5 1I = dx+ dx - dx
16 x - 2 4 16 x+2x - 2
e e5 7 5
= log x - 2 - - log x+2 +C16 4 x - 2 16
Type - 3
When denominator is non-repeated quadratic factors
2 2
f(x) A Bx+CLet = +
x- ax - a px +qx+r px +qx+r
where A, B, C are constants and are determined by either comparing coefficients of similar powers of x or as mentioned in method 1.
Example - 8
2
8Evaluate: dx
x+2 x +4 2
8Solution: Let I = dx
x+2 x +4
22
8 A Bx+CLet = +
x+2 x +4x+2 x +4
28= A x +4 + Bx+C x+2 ... i Identity
Putting x = -2 in i , weget A =1
Putting x = 0 and1 in i , weget
Solution Cont.
C = 2 and B = -1
22
8 1 -x+2= +
x+2 x +4x+2 x +4
21 -x+2
I = dx+ dxx+2 x +4
2 2
1 1 2x 1= dx - dx+2 dx
x+2 2 x +4 x +4 2 -1
e e1 1 x
=log x+2 - log x +4 +2× tan +C2 2 2
2 -1e e
1 x=log x+2 - log x +4 +tan +C
2 2
Type - 4
When denominator is repeated quadratic factors
2 2 2 22 2 2
f(x) Ax+B Cx+D Ex+FLet = + +
ax +bx+c px +qx+rax +bx+c px +qx+r px +qx+r
where A, B, C, D, E and F are constants and are determined by equating the like powers of x on both sides or giving values to x.
Note: If a rational function contains only even powers of x, then we follow the following method:
(i) Substitute x2 = t (ii) Resolve into partial fractions(iii) Replace t by x2
Example – 9
2
2 2
x +2Evaluate: dx
x +1 x +4
2
2 2
x +2Solution: Let I = dx
x +1 x +4
2
2 2
x +2 t+2 A BLet = = +
t+1 t+4 t+1 t+4x +1 x +4
t+4 A+ t+1 Bt+2=
t+1 t+4 t+1 t+4
t 2= t+4 A+ t+1 B Indentity
Putting t =-1, - 4, we get
1 2A = , B =
3 3
Solution Cont.
2
2 2 2 2
t+2 1 2 x +2 1 2= + = +
t+1 t+4 3 t+1 3 t+4 x +1 x +4 3 x +1 3 x +4
2 2
1 1 2 1I = dx+ dx
3 3x +1 x +4
C -1 -11 2 1 x= tan x+ tan
3 3 2 2
C-1 -11 1 x= tan x+ tan
3 3 2
Example - 10
3x dx
Evaluate:x - 1 x - 2
Solution: Here degree of Nr > degree of Dr.
3x 7x - 6
= x+3+x - 1 x - 2 x - 1 x - 2
7x - 6 A B
Let = +x - 1 x - 2 x - 1 x - 2
7x - 6 = x - 2 A+ x- 1 B Identity
Putting x =1, we get A =-1
Solution Cont.
Putting x =2, we get B =8
7x - 6 -1 8
= +x- 1 x - 2 x - 1 x - 2
3x 1 8
= x+3 - + dxx - 1 x - 2 x - 1 x - 2
dx dx= xdx +3 dx - +8
x - 1 x - 2
2
e ex
= +3x - log | x - 1| +8 log | x - 2| +C2
Thank you