Mathematics

Session

Indefinite Integrals -1

Session Objectives

Primitive or Antiderivative

Indefinite Integral

Standard Elementary Integrals

Fundamental Rules of Integration

Methods of Integration 1. Integration by Substitution, Integration Using Trigonometric Identities

Primitive or Antiderivative

then the function F(x) is

called a primitive or an antiderivative of a function f(x).

dIf F x = ƒ x

dx,

5 54 4

For example:

x d xis a primitive of x because = x

5 dx 5

Cont.

If a function f(x) possesses a primitive, then it possesses infinitely many primitives which can be expressed as F(x) + C, where C is an arbitrary constant.

5 5 54

For example:

x x x, - 1, +7 etc., are primitives of x .

5 5 5

Indefinite Integral

Let f(x) be a function. Then collection of all its primitives is called indefinite integral of f(x) and is denoted by

f(x)dx.

d

F x +C = f(x) f(x)dx =F(x) +Cdx

where F(x) + C is primitive of f(x) and C is an arbitrary constant known as ‘constant of integration’.

Cont.

will have infinite number of values and

hence it is called indefinite integral of f(x).

f(x)dx

If one integral of f(x) is F(x), then F(x) + C will be also an integral of f(x), where C is a constant.

d dIf {F(x)}= f(x), then also {F(x)+C}= f(x),

dx dxwhere C is an arbitrary constant.

Standard Elementary Integrals

n+1 n+1

n nd x xi = x x dx = +C, (n -1)

dx n+1 n+1

e ed 1 1

ii log x = dx =log x +C, x 0dx x x

x x

x x

e e

d a aiv =a , a>0, a 1 a dx = +C

dx log a log a

x x x xdiii e =e e dx = e +C

dx

Cont.

dv -cosx = sinx sinxdx =- cosx+C

dx

dvi sinx =cosx cosxdx =sinx+C

dx

2 2dvii tanx =sec x sec xdx =tanx+C

dx

2 2dviii -cotx =cosec x cosec xdx =- cotx+C

dx

dix secx =secxtanx secxtanxdx =secx+C

dx

The following formulas hold in their domain

Cont.

dx -cosecx =cosecxcotx cosecxcotxdx =- cosecx+C

dx

-1 -1

2 2

d 1 dxxi sin x = =sin x+C

dx 1- x 1- x

-1 -1

2 2

d -1 -dxxii cos x = =cos x+C

dx 1- x 1- x

-1 -12 2

d 1 dxviii tan x = =tan x+C

dx 1+x 1+x

Cont.

-1 -12 2

d -1 -dxxiv cot x = =cot x+C

dx 1+x 1+x

-1

2 2

d 1 dxxv sec x = =s

dx x x - 1 x - 1

1 2, x 1 ec x C, x 1x

-1

2 2

d -1 -dxxvi cosec x = =

dx x x - 1 x - 1

1 2, x 1 cosec x C, x 1x

Fundamental Rules of Integration

d1 f x dx = f x

dx

2 kf x dx =k f x dx

3 f(x)±g(x) dx = f x dx± g (x)dx

Example - 1

32

2

1Evaluate: x + dx

x

32

21

Solution: Let I = x + dxx

6 2

6 2

1 3= x + +3x + dx

x x

6 26 2

1 3x dx dx 3x dx dx

x x

6 -6 2 -2= x dx+ x dx+3 x dx+3 x dx

7 -5 3 -1x x 3x x= + + +3× +C

7 -5 3 -1

73

5x 1 3

= - +x - +C7 x5x

Example - 2

1Evaluate: dx

3x+1 - 3x - 1

1Solution: Let I = dx

3x+1 - 3x - 1

1 3x+1+ 3x - 1= × dx

3x+1 - 3x - 1 3x+1+ 3x - 1

3x+1+ 3x - 1

= dx3x+1 - 3x - 1

3x+1+ 3x - 1= dx

2

Cont.

1 1= 3x+1 dx+ 3x - 1 dx

2 2

3 32 2

3x+1 3x - 11 1= + +C

3 32 2×3 ×32 2

3 32 2

1 1= 3x+1 + 3x - 1 +C

9 9

Example - 3

-1Evaluate: cos sinx dx

, x2 2

-1Solution:Let I = cos sinx dx

-1= cos cos - x dx = - x dx2 2

2x= dx - x dx = x - +C

2 2 2

1Note : cos cos 0,

Example - 4

xEvaluate: dx

x+2

xSolution: Let I = dx

x+2x+2- 2

= dxx+2

x+2 1= dx - 2 dx

x+2 x+2

1 1

-2 2= x+2 dx - 2 x+2 dx

3 12 2

32

x+2 x+2= - 2 +C

12

322

= x+2 - 4 x+2 +C3

Integration by Substitution

If g(x) is a differentiable function, then to

evaluate integrals of the form ƒ g x g x dx

We substitute g(x) = t and g’(x) dx = dt,

then the given integral reduced to ƒ t dt

After evaluating this integral, we substitute back the value of t.

Cont.

Integrals of the form of ƒ ax+b dx Let I = ƒ ax+b dx

1Puttingax +b = t adx = dt dx = dt

a

1 1I = ƒ t dt I = F t +C

a a

1I = F ax+b +C, where f x dx =F x +C

a

Example - 5

5Evaluate: (3x+5) dx

5Let I = (3x+5) dx1

Putting 3x+5= t 3dx =dt dx = dt3

5+151 1 t

I= t dt = × +C3 3 5+1

661 1= t +C = 3x+5 +C

18 18

Solution :

Integration Using Trigonometric Identities

m mIntegrals of the form sin nxdx, cos nxdx wherem

is small positive integer can be evaluated using the following

identities.

,

2 1- cos2x1 sin x =

2

2 1+cos2x2 cos x =

2

33 sin3x =3sinx - 4sin x

34 cos3x = 4cos x - 3cosx

Example - 6

4Evaluate: cos x dx4Solution: Let I = cos x dx

2 222cos x 1+cos2x= dx = dx

2 2

21 cos2x cos 2x x sin2x 1+cos4x= + + dx = + + dx

4 2 4 4 4 8

x sin2x 1 1 x sin2x x sin4x= + + dx+ cos4x dx = + + + +C

4 4 8 8 4 4 8 32

3x sin2x sin4x= + + +C

8 4 32

Integration Using Trigonometric Identities

Integrals of the form sinmxcosnxdx, sinmxsinnxdx,

cosmxcosnxdx and cosmxsinnxdx

can be evaluated using the following identities.

2 2cosAsinB = sin A+B - sin A - B

1 2sinAcosB =sin A+B +sin A - B

3 2cosAcosB =cos A+B +cos A - B

4 2sinAsinB =cos A - B - cos A+B

Example - 7

Evaluate: sin3xcos2x dxSolution: Let I = sin3xcos2x dx

1= 2sin3xcos2x dx

2

1= sin5x+sinx dx

2 [Using 2sinAcosB = sin (A + B) + sin (A – B)]

1 cos5xcosx C

2 5

cos5x cosx=- - +C

10 2

Integration by Substitution

ƒ xIntegrals of the form dx

ƒ x

ƒ xLet I = dx

ƒ x

Putting ƒ x = t ƒ ' x dx =dt

e edt

I = =log t +C =log f(x) +Ct

Example - 8

1- tanxEvaluate: dx

1+tanx1- tanx

Solution: Let I = dx1+tanx

cosx - sinx= dx

cosx+sinx

Putting cosx + sinx = t -sinx + cosx dx = dt

edt

I = =log t +Ct

e=log cosx+sinx +C

Solution Cont.

Method - 2

I = tan - x dx4

e= log cos - x C4

Example - 9

x

2 x

x+1 eEvaluate: dx

cos xe

x

2 x

x+1 eSolution: Let I = dx

cos xe

x x x xPutting xe = t xe + e dx = dt x +1 e dx = dt

2 xI = sec t dt = tant+C = tan xe +C

Some Standard Results

e e1 tanxdx =-log cosx +C =log secx +C

e2 cotxdx =log sinx +C

e ex

3 secx dx =log secx+tanx +C =log tan + +C4 2

e ex

4 cosecxdx =log cosecx - cotx +C =log tan +C2

Integration by Substitution

nIntegrals of the form ƒ x f' x dx

nLet I = ƒ x f' x dx

Puttingƒ x = t ƒ ' x dx =dt

n+1n+1n f xt

I = t dt I = +C I = +C, n -1n+1 n+1

Example - 10

3elog xEvaluate: dx

x 3elog x

Solution: Let I = dxx

e1

Putting log x = t dx =dtx

3I = t dt 44

elog xt= +C = +C

4 4

Integration by Substitution

m nIntegrals of the form sin xcos xdxUse the following substitutions.(i) When power of sinx i.e. m is odd, put cos x = t,(ii) When power of cosx i.e. n is odd, put sinx = t,(iii) When m and n are both odd, put either sinx = t or cosx = t,(iv) When both m and n are even, use De’ Moivre’s theorem.

Example - 11

3 5Evaluate: sin xcos xdx3 5Solution: Let I = sin xcos xdx

Powers of sin x and cos x are odd.

Therefore, substitute sinx = t or cosx = t

1Substituting cosx = t -sinxdx =dt dx =- dt

sinx

3 5 1I = sin xt - dt

sinx

We should put cosx = t, because power of cosx is heigher

Cont.

5 2=- t sin xdt 5 2=- t 1- t dt

5 7=- t - t dt6 8 6 8t t cos x cos x

=- + +C =- + +C6 8 6 8

Example - 12

-1

2

sin 2tan xEvaluate: dx

1+x

-1

2

sin 2tan xSolution: Let I = dx

1+x

-12

1Putting tan x = t dx =dt

1+x

I = sin2t dt cos2t

=- +C2

-1cos2 tan x=- +C

2

Example - 13

xEvaluate: e - 1 dxxSolution: Let I = e - 1 dx

x 2 xx 2

2t dt 2t dtPutting e - 1 = t e dt = 2tdt dx = dx =

e 1+ t

22

2 22t dt t dt

I = t . =21+t 1+t

2

2 2t +1- 1 1

=2 dt =2 dt - dt1+t 1+t

-1 x -1 x= 2t - 2tan t +C = 2 e - 1 - 2tan e - 1 +C

Example - 14

x x55 5 xEvaluate : 5 5 5 dx

x x x5 5 35 5 5 x

eSubstituting 5 = t 5 5 5 log 5 dx = dt

x x55 5 x

3

e

15 5 5 dx = dt

log 5

x x55 5 xSolution : Let I = 5 5 5 dx

3 3

e e

1 1I = dt = t +C

log 5 log 5

x55

3

e

5= +C

log 5

Thank you