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