Maths Integration Notes : MHT-CET

1

Indefinite IntegralsIndefinite IntegralsIndefinite IntegralsIndefinite Integrals

Indefinite Integrals Definition Indefinite Integrals Definition Indefinite Integrals Definition Indefinite Integrals Definition

If d

dx[ g (x) ] = f (x), then g (x) + c is an indefinite integral of f (x) and

we write it as ( )f x dx∫ = g (x) + c.

e.g.

2

2

d( x ) 2x

dx

2x dx x c

=

∴ = +∫

ImportanImportanImportanImportant t t t

The results which are true for x are true for ax + b also ( as both are

linear ), the only change is, obtained answer is to be divided by a ( i.e.

coefficient of x )

For example For example For example For example

2sec x dx tan x c= +∫

∴ 2 tan (x / 2)x x

sec dx c 2 tan c2 1/ 2 2

= + = +

∫

2

Different typesDifferent typesDifferent typesDifferent types

Type Type Type Type –––– I I I I

Sums based on standard result.

Type Type Type Type –––– IIIIIIII

Integrand of the type P (x)

ax b+ , where P (x) is a polynomial

Here divide numerator by denominator and use

log |ax b |dxc

ax b a

+= +

+∫

Type Type Type Type ---- IIIIIIIIIIII

Integration by substitutionIntegration by substitutionIntegration by substitutionIntegration by substitution

'( )log ( )

( )

f xdx f x c

f x= +∫

Standard SubstitutionsStandard SubstitutionsStandard SubstitutionsStandard Substitutions

Expression Substitution 2 2a x− x = a sin θ or x = a cos θ 2 2

a x+ x = a tan θ or x = a cot θ 2 2

x a− x = a sec θ or x = a cosec θ

a x

a x

−

+ x = a cos θ

3

Type Type Type Type –––– IVIVIVIV

Integrals of the type

x

x

a sin x b cos x a e bdx and dx

csin x d cos x ce d

+ +

+ +∫ ∫

Here we write

dNumerator A ( Denominator ) B ( Denominator)

dx= +

Values of A and B are to be obtained by equating coefficients of sin x

and cos x.

Type Type Type Type –––– VVVV

To express angle in numerator in terms of angle in the denominator.

e.g.

To evaluate sin ( x a )

dxsin ( x a )

−

+∫

We write

x – a = ( x + a ) – 2a

Type Type Type Type –––– VIVIVIVI

Integral of type

2 22

px q px q P(x )dx OR dx OR dx

ax bx c ax bx cax bx c

+ +

+ + + ++ +∫ ∫ ∫

Here we write

2dpx q A B ( ax bx c )

dx+ = + + +

In the third integral P(x) is a polynomial of degree greater than or

equal to 2. Here, divide numerator by denominator and then proceed.

4

Note Note Note Note : In CET, values of A and B can be obtained directly.

e.g. in 2

4x 3dx

x 2x 1

+

+ +∫

As derivative of x 2 + 2x + 1 is 2x + 2

we write 4x + 3 = 2 ( 2x + 2 ) – 4 + 3 = 2 ( 2x + 2 ) – 1

Type Type Type Type –––– VIIVIIVIIVII ( Important )( Important )( Important )( Important )

Integral of the type 1 1 1

dx OR dx OR dxa b sin x a b cos x a b sin x ccos x+ + + +∫ ∫ ∫

Here we put x

tan t2

=

2

2 2 2

2dt 1 t2tdx , sin x , cos x

1 t 1 t 1 t

−∴ = = =

+ + +

Note :Note :Note :Note :

If angle is 2x, put tan x = t 2

2 2 2

1 tdt 2tdx , sin x , cos x

1 t 1 t 1 t

−∴ = = =

+ + +

5

Type Type Type Type –––– VIIVIIVIIVIIIIII

Integral of the type

2 2 2 2

1 1 1dx OR dx OR dx

a b sin x a b cos x a b sin x c cos x+ + + +∫ ∫ ∫

Here we multiply numerator and denominator by sec 2 x and put

tan x = t

Note :Note :Note :Note :

In the denominator

sin 2 x × sec 2 x = tan 2 x

cos 2 x × sec 2 x = 1

sec 2 x = 1 + tan 2 x

Type Type Type Type –––– IXIXIXIX

Integration By partsIntegration By partsIntegration By partsIntegration By parts

The Theorem is

duuv dx u v dx v dx dx

dx

= −

∫ ∫ ∫ ∫

This can be remembered as

fis – idfis Rule

The order in which u and v are to be taken is according to the serial

order of the letters of the word “LIATE”, where

L : Logarithmic

I : Inverse Trigonometric

A : Algebraic

T : Trigonometric

E : Exponential

6

Type Type Type Type –––– XXXX

Sums on

1. 2 2sin ( ) ( sin cos )

a xa x e

e bx dx a bx b bx ca b

= − ++∫

2. 2 2cos ( ) ( cos sin )

a x

a x ee bx dx a bx b bx c

a b= + +

+∫

Type Type Type Type –––– XIXIXIXI

Sums based on x xe [f (x) f '(x)] dx e f (x) c+ = +∫

Here multiple of e x is expressed as sum of a function and its

derivative.

Typical sums of this type 2

1tan x

2 2

log x 1 x xdx , e dx

(1 log x ) 1 x

− + +

+ +∫ ∫

7

Type Type Type Type –––– XIIXIIXIIXII

Partial FractionsPartial FractionsPartial FractionsPartial Fractions

1. 1. 1. 1. Distinct Linear FactorsDistinct Linear FactorsDistinct Linear FactorsDistinct Linear Factors

ExampleExampleExampleExample

2

3x 5 3x 5 A B

( x 1) ( x 3) ( x 1) ( x 3)x 2x 3

+ += = +

− + − ++ −

We write

3x + 5 = A ( x + 3 ) + B ( x – 1 )

And so on

Disguised linear factorsDisguised linear factorsDisguised linear factorsDisguised linear factors

Example Example Example Example

In

2

2 2

x

( x 4) ( x 7 )+ − we take x 2 = t for finding partial fractions

only. This is not a substitution.

2. 2. 2. 2. Repeated Linear FactorsRepeated Linear FactorsRepeated Linear FactorsRepeated Linear Factors

ExampleExampleExampleExample

2 2

x 1 A B C

( x 1) ( x 2)( x 1) ( x 2 ) ( x 1)

+= + +

− +− + −

8

3.3.3.3. Non Non Non Non –––– repeated repeated repeated repeated quadraticquadraticquadraticquadratic factorfactorfactorfactor

Example Example Example Example

2 2

Bx cx A

( 2x 1)( 2x 1) ( x 3) ( x 3)

+= +

++ + +

Type Type Type Type –––– XIIIXIIIXIIIXIII

Reduction FormulaeReduction FormulaeReduction FormulaeReduction Formulae

1. 1 21 1

sin sin cos sinn n nnx dx x x x dx

n n

− −− −= +∫ ∫

2. 1 21 1

cos cos sin cosn n nnx dx x x x dx

n n

− −−= +∫ ∫

Poll QuestionPoll QuestionPoll QuestionPoll Question

log f (x)e f (x)=

9

Proof :Proof :Proof :Proof :

y

e

y

log f (x)e

Let e f (x)

y log f (x)

Put this in e f (x) to get

e f (x)

=

∴ =

=

=

Solved SumsSolved SumsSolved SumsSolved Sums

CET CET CET CET –––– 2008 ( Memory Based )2008 ( Memory Based )2008 ( Memory Based )2008 ( Memory Based )

1. 2 2

3 dx

( x 1) ( x 4)=

+ +∫

(a) log ( x 2 + 1 ) – log ( x 2 + 4 ) + c

(b) 1 11 x

tan x tan c2 2

− − − +

(c) 1 1 x

2 tan x tan c2

− − − +

(d) 1 11

tan x tan x c2

− −− +

SolutionSolutionSolutionSolution

2 2 2 2

1 1

3 dx 1 1I dx

( x 1) ( x 4) x 1 x 4

1 xtan x tan c

2 2

− −

= = −

+ + + +

= − +

∫ ∫

2. 2

1 1dx

log x ( log x )

− =

∫

10

(a) 1

clog x

+ (b) x

clog x

+ (c) ( ) 2

xc

log x+ (d) log x + c

Solution Solution Solution Solution

2

1 1I dx

log x ( log x )

= −

∫

Put log x = t ∴ x = e t ∴ dx = e t dt

t t

2

1 1 1 xI e dt e c c

t t log xt

∴ = − = + = +

∫

3.

x

3

x edx

( x 2)=

+∫

(a)

xec

x 2+

+ (b)

( )

x

2

ec

x 2+

+

(c) ( )

x

3

ec

x 2+

+ (d)

( )

x

2

x ec

x 2+

+


( )

x xx

3 3 2 3

x

2

x e e ( x 2 2) 1 2I dx dx e dx

( x 2) ( x 2) ( x 2) ( x 2)

ec

x 2

+ −= = = −

+ + + +

= ++

∫ ∫ ∫

4. 2 2 2

1dx

a b x=

−∫

(a) 11 bx

sin cb a

− +

(b)

11 axsin c

ab b

− +

(c) 11 ax

sin cb b

− +

(d)

11 bxsin c

ab a

− +

11


2 2 2 2

2

1

1 1 1I dx dx

ba b x ax

b

1 bxsin c

b a

−

= =−

−

= +

∫ ∫

5. n

1dx

x x −=

+∫

(a) n 11

log | x 1| cn 1

++ +

+ (b)

n1log | x | c

n+

(c) n1

log | x | cn 1

++

(d) n 11

log | x | cn

++


n

n n 1

n

n

n 1

n 1

x1 1I dx dx dx

1x x x 1x

x

( n 1 ) x1dx

n 1 x 1

1log | x 1 | c

n 1

− +

+

+

= = =+ +

+

+=

+ +

= + ++

∫ ∫ ∫

∫

CET CET CET CET –––– 2009 ( Memory Based)2009 ( Memory Based)2009 ( Memory Based)2009 ( Memory Based)

1. x

2

x 1e dx

x

−=

∫

(a)

x

2

ec

x+ (b)

xec

x+ (c)

xx e c+ (d) x 2e x c+

12


x x

2

1 1 1I e dx e c

x xx

= − = × +

∫

2. x log x dx =∫

(a) ( )x

2 log x 1 c4

− + (b) ( )2x

2 log x 1 c4

− +

(c) ( )2x

2 log x 1 c4

+ + (d) ( )x

2 log x 1 c4

+ +


( )

2 2

2 2 2

I x log x dx log x . x dx

x x1log x

2 x 2

x x x1log x 2 log x 1 c

2 2 2 4

= =

= × − ×

= − × = − +

∫ ∫

∫

CET CET CET CET –––– 2010 ( Memory Based )2010 ( Memory Based )2010 ( Memory Based )2010 ( Memory Based )

1. 2

dx

16 x 9=

+∫

(a) 11 4x

tan c4 3

− +

(b)

11 4xtan c

12 3

− +

(c) 11 4x

tan c3 3

− +

(d)

1 4xtan c

3

− +

SolSolSolSolutionutionutionution

13

22

1

1

dx 1 dxI

91616 x 9x

16

11 xtan c

316 3/ 4

4

1 4xtan c

12 3

−

−

= =+

+

= × +

= +

∫ ∫

2. tan x 2 3e (sec x sec x sin x ) dx+ =∫

(a) e x sec x tan x + c (b) e tan x . tan x + c

(c) e tan x sec 2 x + c (d) e tan x . tan 2 x + c


tan x 2 3 tan x 2

tan x 2

2

x t tan x

e (sec x sec x sin x ) dx e sec x ( 1 sec x sin x ) dx

e ( 1 tan x)sec x dx

put tan x t sec x dx dt

I e ( t 1) dt e . t e . tan x c

+ = +

= +

= ∴ =

∴ = + = = +

∫ ∫

∫

∫

More Solved SumsMore Solved SumsMore Solved SumsMore Solved Sums

1.

5 log 4 log

3 log 2 log

x x

x x

e edx

e e

−

−∫

(a) log | x 3 – x 2 | + c (b)

3

3

xc+

(c) 2

xc+ (d) log | x ( x – 1 ) | + c

14


( ) ( )

( )

( )

x

45 4

3 2 2

32

5 4

3 2

log x log x

log x log x

log f

e eI

e e

using e f x

x x 1x xI

x x x x 1

xx dx

3

-=

-

=

--= =

- -

= =

ò

ò ò

ò

2. cos 2 cos 2

cos cos

xdx

x

α

α

−

−∫ is

(a) sin x – x sin α + c (b) x cos α + cos x + c

(c) 2 ( sin x + x cos α ) + c (d) cos x + x sin α + c


( )

( )

( ) ( )

( ) ( ) ( )

2 2

2 2

2cos x 1 2cos 1I dx

cos x cos

2 cos x cosdx

cos x cos

cos x cos cos x cos2

cos x cos

2 cos x cos dx 2 sin x x cos c 2 sin x x cos c

− − α −=

− α

− α=

− α

+ α − α=

− α

= + α = + α + = + α +

∫

∫

∫

∫

3. 1 2 tan (sec tan )x x x dx+ +∫ is

(a) 3/23

[1 2 tan (sec tan ) ]2

x x x c+ + +

(b) log Gsec x ( sec x + tan x ) G + c

(c) log Gsec x G - log G sec x + tan x G + c

(d) log G sec x + tan x G + c

15


( )

( ) ( )

[ ] [ ]

( )

2

2 2

2 2

2

I 1 2 tan x sec x 2 tan x

1 tan x 2sec x tan x tan x

sec x 2sec x tan x 2 tan x dx

sec x tan x dx sec x tan x dx

log sec x tan x log sec x c

log sec x sec x tan x 1 c

= + +

= + + +

= + +

= + = +

= + + +

= + +

∫

∫

∫

∫ ∫

4. 4 4

sin 2

cos sin

x dx

x x+∫ =

(a)

4

4

coslog

sin

x

x + c (b) cot – 1 ( tan x ) + c

(c) tan – 1 ( tan 2 x ) + c (d) 2 tan – 1 ( tan x ) + c


( )

( ) ( )

4 4

4

2 2

4 4

2

22

2 2

1 1 2

2

2sin x cos xI dx

cos x sin x

Divide Numerator and Deno min ator by cos x

sin x 12 dx

cos x cos x 2 tan x sec x dx

1 tan x tan x 1

2 tan x sec x dxI

tan x 1

put tan x t 2 tan x sec x dx dt

dtI tan t c tan tan x c

t 1

− −

=+

= =+ +

=

+

= ∴ =

= = + = ++

∫

∫ ∫

∫

∫

5. cos sin

1 sin 2

x xdx

x

−

+∫

16

(a) 1

cos sinc

x x

−+

+ (b)

1

cos sinc

x x+

−

(c) 1

sin 2c

x+ (d)

1

cos 2c

x+


( )

( )

( )2 2 2

2

2

2

1

put cos x sin x t ........ 1

sin x cos x dx dt

on squaring 1

cos x sin x 2 sin x cos x t

1 sin 2 x t

dtI t dt

t

t 1I c

1 cos x sin x

−

−

+ =

∴ − + =

+ + =

∴ + =

= =

−= = +

− +

∫ ∫

6. log( 2) log

( 2)

x xdx

x x

+ −

+∫

(a)

2

1 2log

4

xc

x

++

(b) [log(x + 2) – log x ] 2 + c

(c)

2

1 2log

4

xc

x

− ++

(d) [log(x + 2) – log x ] + c


17

( )

( )

( )2

2

2

put log x 2 log x dt

1 1d x dt

x 2 x

x x 2dx dt

x x 2

dx 1dt

x x 2 2

t1 1I t dt

2 2 2

1log x 2 log x c

4

x 21log c

4 x

+ − =

∴ − =

+

− −∴ =

+

−∴ =

+

− −∴ = × = ×

−= + − +

+ −= +

∫

7. cos( )cos( )

dx

x a x b− −∫ =

(a) log Gcos ( x – a ) cos ( x – b ) G+c

(b) tan( )

logtan( )

x ac

x b

−+

−

(c) 1 cos( )

logsin( ) cos( )

x ac

a b x b

−+

− −

(d) 1 cos( )

logcos( ) cos( )

x ac

a b x b

−+

− −


18

( )

( ) ( )

( ) ( )

( )

( ) ( ) ( ) ( )

( ) ( )

( )( ) ( )( )

( )( ) ( )

( )

( )

( )

( )

( )

( )

sin x b x a1I

sin a b cos x a cos x b

sin x b cos x a cos x b sin x a1

sin a b cos x a cos x b

1tan x b tan x a dx

sin a b

1log sec x b log sec x a

sin a b

sec x b1log c

sin a b sec x a

cos x a1log c

sin a b cos x b

− − − =

− − −

− − − − −=

− − −

= − − −−

= − − − −

−= +

− −

−= +

− −

∫

∫

∫

8. ( 1) 1

dx

x x+ −∫ =

(a) 1 1

2 tan2

xc−

−+

(b)

11 1tan

22

xc

− +

+

(c) 1 1

tan2

xc

− −

+

(d) 11 1

tan2 2

− +

+

xc


( )

2

2

22

1 1

put x 1 t dx 2 tdt

also x 1 t

2 t d t d tI 2

t 21 t 1 t

x 11 t2 tan c 2 tan c

2 2 2

− −

− = ∴ =

= +

= =++ +

− = × + = +

∫ ∫

9. 4 5sin

dx

x−∫

19

(a) 13 5 tan( / 2) 4

tan2 3

xc

− − +

(b)

( )

( )

2 tan / 2 41log

3 2 tan / 2 1

−

−

x

x

(c) log 5 4sinx c− + (d) 2

log 5 tan / 2 43

x c+ +


2 2

2

2

2

2 22

xput tan t

2

2 dt 2tdx sin x

1 t 1 t

2dt

1 t dtI 2

2t 4 4 t 10 t4 51 t

2 dt 1 dt 1 dt

5 t4 2 25 25 5 9t 1 t 1 t2 4 16 4 16

5 3t

1 1 4 4log3 5 32

2 t4 4 4

t 2 2 t 4 2 tan (x / 2) 41 1 1log log log

13 3 2 t 1 3t

2

=

∴ = =+ +

+= =

+ −−+

= = =

− + − − + − −

− −

= ×

× − +

− − −= = =

−−

∫ ∫

∫ ∫ ∫

2 tan (x / 2) 1−

10. cos

(1 sin )(2 sin )

x dx

x x− −∫

(a) 1 sin

log2 sin

xc

x

−+

− (b) log (1 sin )(2 sin )x x c− − +

(c) 2 sin

log1 sin

xc

x

−+

− (d)

(1 sin )log

(2 sin )

−+

−

xc

x

SoluSoluSoluSolutiontiontiontion

20

( ) ( )

( ) ( )

( ) ( )

put sin x t cos x dx dt

2 t 1 tdtI

1 t 2 t 1 t 2 t

1 1dt

1 t 2 t

log 1 t log 2 tI

1 1

2 t 2 sin xI log c log c

1 t 1 sin x

= ∴ =

− − −= =

− − − −

= −

− −

− −∴ = −

− −

− −= + = +

− −

∫ ∫

∫

11.

11 12 13

14 15

x x xdx

x x

+ +

+∫

(a) 14 15log x x c+ + (b) 2

1

1 2

xlog c

x x− +

+

(c) 2

1

1 2

xlog c

x x+ +

+ (d) 2

1

1− +

+

xlog c

x x


21

( )

( )

( )

( ) ( )

( )

( )

11 12 13 11 2

14 15 14

22

3 3

3

3

2

2

2

x x x x (1 x x )I dx dx

x x x (1 x )

1 x x1 x x

x 1 x x 1 x

1 x 1

x 1 xx 1 x

1 x xx dx dx

x 1 x

x 1 1dx

2 x 1 x

1log x log 1 x c

2 x

x1log c

1 x2 x

−

−

+ + + += =

+ +

+ ++ += =

+ +

+= +

++

+ −= +

+

= + −

− +

−= + − + +

−= + +

+

∫ ∫

∫ ∫

∫ ∫

∫ ∫

∫

12. cosxe x dx∫

(a) 1

(sin 2cos )2

xe x x c+ + (b)

1(sin cos )

2

xe x x c− +

(c) 1

(cos sin )2

xe x x c− + (d)

1(sin cos )

2

xe x x c+ +


[ ]

[ ]

x

axa x

2 2

x

I e .cos x .dx

eusing e cos bx dx a cos bx b sin bx c

a b

eI cos x sin x c

1 1

=

= + ++

= + ++

∫

∫

Maths Integration Notes : MHT-CET

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