( u - deepaksirmaths.weebly.com fileMATHEMATICS 383 Notes MODULE - V Calculus Integration I function...

MATHEMATICS

Notes

MODULE - VCalculus

382

Integration

or ( ) ( ) ( )d d df g fg g f

dx dx dx= − (1)

Also you know that ( )dfg dx fg

dx=∫

Integrating (1), we have

( ) ( ) ( )d d df g dx fg dx g f dx

dx dx dx= −∫ ∫ ∫

( )dfg g f dx

dx= − ∫ (2)

if we take ( ) ( ) ( )df u x ; g v x ,

dx= =

(2) becomes ( ) ( )u x v x dx∫

( ) ( ) ( )( ) ( )du x v x dx u x v x dx dx

dx = ⋅ − ∫ ∫ ∫

= I function × integral of II function −∫ [differential coefficient of I function × integralof II function] dx

A B

Here the important factor is the choice of I and II function in the product of two functionsbecause either can be I or II function. For that the indicator will be part 'B' of the result above.The first function is to be chosen such that it reduces to a next lower term or to a constant termafter subsequent differentiations.

In questions of integration like2 2 xxsinx, x cos x, x e

(i) algebraic function should be taken as the first function

(ii) If there is no algebraic function then look for a function which simplifies the product in 'B'as above; the choice can be in order of preference like choosing first function

(i) an inverse function (ii) a logarithmic function

(iii) a trigonometric function (iv) an exponential functionThe following examples will give a practice to the concept of choosing first function.

I function II function

1. x cosxdx∫ x (being algebraic) cosx

2. 2 xx e dx∫ 2x (being algebraic) xe

3. 2x log x dx∫ log x 2x

MATHEMATICS 383

Notes

MODULE - VCalculus

Integration

I function II function

4. ( )2log x

dx1 x+∫ log x ( )2

1

1 x+

5. 1xsin x dx−∫ 1sin x− x

6. logxdx∫ log x 1

(In single function oflogarithm and inversetrigonometric we takeunity as II function)

7. 1sin xdx− 1sin x− 1

Example 26.37 Evaluate :

x cos x dx∫Solution : Taking the polynomial (algebraic function) x as the first function and trigonometricfunction cos x as the second function, we get

x cos x dx

I II∫ ( )d

x cos x dx x cosxdx dxdx

= − ⋅ ∫ ∫ ∫

Ifunction × Integral of II function −

( ) ( )dI function II function dx dx

dx ⋅ ∫ ∫

xs inx 1.sinx dx= − ∫[ ]xs inx cosx C= − − +

xsinx cosx C= + +

Example 26.38 Evaluate :2x sinxdx∫

Solution : Taking algebraic function 2x as I function and sin x as II function, we have,

2x sinxdxI II∫ ( )2 2d

x sin x x sinx dx dxdx

= − ∫ ∫ ∫

( )2x cosx 2 x cosx dx= − − −∫ 2x cosx 2 x cosx dx= − + ∫ (1)

Again, x cos x dx xs inx cosx C= + +∫ (2)

Substituting (2) in (1), we have

MATHEMATICS

Notes

MODULE - VCalculus

384

Integration

[ ]2 2x sin x dx x cosx 2 xs inx cosx C= − + + +∫ 2x cosx 2xsinx 2cosx C= − + + +

Example 26.39 Evaluate :2x log x dx∫

Solution : In order of preference log x is to be taken as I function.

∴ 2log x x dx

I II∫

3 3x 1 xlog x dx

3 x 3= − ⋅∫

3 2x xlog x dx

3 3= − ∫

3 3x 1 xlog x C

3 3 3

= − +

3 3x xlog x C

3 9= − +


( )2log x

dx1 x+∫

Solution : ( )2

log xdx

1 x+∫ ( )21

log x dx1 x

I II

= ⋅+∫

1 1 1log x dx

1 x x 1 x− = − − × + + ∫

( )log x 1

dx1 x x 1 x

−= ++ +∫

log x 1 1dx

1 x x 1 x− = + − + + ∫

log x 1 1dx dx

1 x x 1 x−= + −

+ +∫ ∫log x

log x log 1 x C1 x

−= + − + ++

log x xlog C

1 x 1 x−

= + ++ +

MATHEMATICS 385

Notes

MODULE - VCalculus

Integration

Example 26.41 Evaluate :2xx e dx∫

Solution : 2xx e dx∫2x 2xe e

x 1. dx2 2

= − ∫2x 2xe 1 e

x C2 2 2

= − +

2x2xe 1

x e C2 4

= − + .


1sin x dx−∫Solution : 1sin x dx−∫ 1sin x 1 dx−= ⋅ ⋅∫

12

xxsin x dx

1 x−= −

−∫

Let 21 x t− =⇒ 2x dx dt− =

⇒ 1

x dx dt2−=

∴ 2

x 1 dtdx

2 t1 x= −

−∫ ∫

t C= − +

21 x C= − − +

∴ 1 1 2sin x dx x sin x 1 x C− −= + − +∫

CHECK YOUR PROGRESS 26.6

Evaluate :

1. (a) xs inx dx∫ (b) ( )21 x cos2x dx+∫ (c) xsin2xdx∫2. (a) 2xtan xdx∫ (b) 2 2x sin xdx∫3. (a) 3x log2xdx∫ (b) ( )21 x logxdx− (c) ( )2log x dx∫

4. (a) nlog x

dxx∫ (b)

( )log log xdx

x∫

MATHEMATICS

Notes

MODULE - VCalculus

386

Integration

5. (a) 2 3xx e dx∫ (b) 3xx e dx∫6. ( )2x log x dx∫7. (a) 1sec xdx−∫ (b) 1x cot xdx−∫

26.7 INTEGRAL OF THE FORM ( ) ( )[ ]xe f x f ' x dx+∫( ) ( )[ ]xe f x f ' x dx+∫

where f ' (x) is the differentiation of f (x). In such type of integration while integrating by parts the

solution will be ( )( )xe f x C.+

For example, consider

[ ]xe tanx logsecx dx+∫ (1)

Let ( )f x logsecx= ,

then ( ) sec x tan xf ' x tan x

sec x= =

So (1) can be rewritten as

( ) ( )[ ] ( )( )x x xe f ' x f x dx e f x C e logsecx C+ = + = +∫Alternatively, you can evaluate it as under :

[ ]x x xe tanx logsecx dx e tanxdx e logsecx dx

I II

+ = +∫ ∫ ∫

x x xe logsecx e logsecx dx e logsecx dx= − +∫ ∫xe logsecx C= +

Example 26.43 Evaluate the following :

(a) x2

1 1e dx

x x − ∫ (b) x 1 x logx

e dxx

+ ∫

(c)( )

x

2x e

dxx 1+∫ (d) x 1 s i n x

e dx1 cosx

+ + ∫

Solution :

(a)x x

21 1 1 d 1

e dx e dxx x dx xx

− = + ∫ ∫ x 1e

x =

(b)x x1 x log x 1

e dx e logx dxx x

+ = + ∫ ∫

MATHEMATICS 387

Notes

MODULE - VCalculus

Integration

( )x de log x log x dx

dx = + ∫

xe log x C= +

(c) ( ) ( )

xx

2 2x e x 1 1

dx e dxx 1 x 1

+ −=+ +∫ ∫

( )x

21 1

e dxx 1 x 1

= − + +

∫

( )x 1 d 1

e dxx 1 dx x 1

= + + +

∫

x 1e C

x 1 = + +

(d) x x

2

x x1 2sin cos1 s i n x 2 2e dx e dx

x1 cosx 2cos2

+ + = +

∫ ∫

x 21 x x

e sec tan dx2 2 2

= + ∫

x x d x

e tan tan dx2 x 2

= + ∫

x xe tan C

2= +

Example 26.44 Evaluate the following :

(a) 3sec x dx∫ (b) xe sinx dx∫Solution :

(a) 3sec x dx∫Let 2I sec x sec x dx= ⋅∫

secx tan x sec x tan x tan x dx= ⋅ − ⋅∫∴ ( )3I secxtanx sec x secx dx= − −∫ ( )2 2tan x sec x 1= −∵

or 3I secxtanx sec x dx sec x dx= − +∫ ∫or 2I secxtanx secx dx= + ∫

MATHEMATICS

Notes

MODULE - VCalculus

388

Integration

or 1I secxtanx log sec x tanx C= + + +

or [ ]1I sec x tan x log sec x tan x C

2= + + +

(b) xe sinx dx∫Let xI e sinx dx= ∫

( ) ( )x xe cosx e cosx dx= − − −∫ x xe cosx e cosx dx= − + ∫ ( )x x xe cosx e sin x e sinx dx= − + − ∫

∴ x xI e cosx e sinx I= − + −or x x2I e cosx e s i n x= − +

or ( )xe

I sinx cosx C2

= − +


2 2a x dx−∫Solution :

Let 2 2 2 2I a x dx a x .1dx= − = −∫ ∫Integrating by parts only and taking 1 as the second function, we have

( ) ( )2 22 2

1I a x x 2x x dx

2 a x= − − − ⋅

−∫

2

2 22 2

xx a x dx

a x= − +

−∫

( )2 2 2

2 22 2

a a xx a x dx

a x

− −= − +

−∫

2 2 2 2 2

2 2

1x a x a dx a x dx

a x= − + − −

−∫ ∫

∴2 2 2 1 x

I x a x a sin Ia

− = − + −

or2 2 2 1 x

2I x a x a sina

− = − +

MATHEMATICS 389

Notes

MODULE - VCalculus

Integration

or2 2 2 11 x

I x a x a sin C2 a

− = − + +

Similarly,2 2 2

2 2 2 2x x a ax a dx log x x a C

2 2−− = − + − +∫

∴2 2 2

2 2 2 2x a x aa x dx log x a x C

2 2++ = − + + +∫


(a) 216x 25dx+∫ (b) 216 x dx−∫ (c) 21 x 2x dx+ −∫Solution :

(a)2

2 2 225 516x 25dx 4 x dx 4 x dx

16 4 + = + = + ∫ ∫ ∫

Using the formula for ( )2 2x a dx+∫ we get,

2 2 2x 25 25 2516x 25dx x log x x C

2 16 32 16

+ = + + + + +

∫

2 2x 2516x 25 log 4x 16x 25 C

8 8= + + + + +

(b) Using the formula for ( )2 2a x dx−∫ we get,

( )22 216 x dx 4 x dx− = −∫ ∫2 1x 16 x

16 x sin C2 2 4

−= − + +

(c) 2 21 x1 x 2x dx 2 x dx

2 2+ − = + −∫ ∫

21 x 1 12 x dx

2 2 16 16 = − − + + ∫

2 23 12 x dx

4 4 = − − ∫

21

1 1x x9 1 94 42 x sin C

32 16 4 16 24

− − − = − − + + ×

MATHEMATICS

Notes

MODULE - VCalculus

390

Integration

2 14x 1 1 9 4x 12 1 x 2x sin C

8 32 32−− − = ⋅ + − + +

2 14x 1 9 2 4x 11 x 2x sin C

8 32 3−− −= + − + +


Evaluate :

1. (a) [ ]xe secx 1 tan x dx+∫ (b) [ ]xe secx log sec x tan x dx+ +∫2. (a)

x2

x 1e dx

x

−∫ (b) x 1

2

1e sin x dx

1 x−

− −

∫

3.( )( )

x3

x 1e dx

x 1

−

+∫ 4. ( )

x

2xe

dxx 1+∫

5.x s i n x

dx1 cosx

++∫ 6. xe sin2x dx∫

26.8 INTEGRATION BY USING PARTIAL FRACTIONS

By now we are equipped with the various techniques of integration.

But there still may be a case like 24x 5

x x 6

++ −

, where the substitution or the integration by parts

may not be of much help. In this case, we take the help of another technique called techniqueof integration using partial functions.

Any proper rational fraction ( )( )

p xq x can be expressed as the sum of rational functions, each

having a single factor of q(x). Each such fraction is known as partial fraction and the processof obtaining them is called decomposition or resolving of the given fraction into partial fractions.

For example, ( ) ( ) 23 5 8x 7 8x 7

x 2 x 1 x 2 x 1 x x 2+ ++ = =

+ − + − + −

Here 3 5

,x 2 x 1+ −

are called partial fractions of 28x 7

x x 2

++ −

.

If ( )( )

f xg x is a proper fraction and g (x) can be resolved into real factors then,

(a) corresponding to each non repeated linear factor ax + b, there is a partial fraction of theform

Aax b+

MATHEMATICS 391

Notes

MODULE - VCalculus

Integration

(b) for ( )2ax b+ we take the sum of two partial fractions as

( )2A B

ax b ax b+

+ +

For ( )3ax b+ we take the sum of three partial fractions as

( ) ( )2 3A B C

ax b ax b ax b+ +

+ + +and so on.

(c) For a non-fractorisable quadratic polynomial 2ax bx c+ + there is a partial fraction

2Ax B

ax bx c

++ +

Therefore, if g (x) is a proper fraction ( )( )

f xg x and can be resolved into real factors, then

( )( )

f xg x

can be written in the following form :

Factor in the denominator Corresponding partial fraction

ax b+A

ax b+

( )2ax b+ ( ) ( )2A B

ax b ax b+

+ +

( )3ax b+ ( ) ( )2 3A B C

ax b ax b ax b+ +

+ + +

2ax bx c+ + 2Ax B

ax bx c

++ +

( )22ax bx c+ + ( )2 22

Ax B Cx Dax bx c ax bx c

+ +++ + + +

where A, B, C, D are arbitrary constants.The rational functions which we shall consider for integration will be those whose denominatorscan be fracted into linear and quadratic factors.


22x 5

dxx x 2

+− −∫

MATHEMATICS

Notes

MODULE - VCalculus

392

Integration

Solution : ( )( )22x 5 2x 5

x 2 x 1x x 2+ +=

− +− −

Let ( )( )2x 5 A B

x 2 x 1 x 2 x 1+ = +

− + − + (1)

Multiplying both sides by ( ) ( )x 2 x 1− + , we have

( ) ( )2x 5 A x 1 B x 2+ = + + −

Putting x = 2, we get 9 = 3A or A = 3

Putting x 1= − , we get 3 3B= − or B 1= −Substituting these values in (1), we have

( )( )2x 5 3 1

x 2 x 1 x 2 x 1+ = −

− + − +

⇒ 22x 5 3 1

dx dx dxx 2 x 1x x 2

+= −

− +− −∫ ∫ ∫ 3log x 2 log x 1 C= − − + +


(a)2

3 2x x 1

dxx x 6x

− −− −∫ (b) ( ) ( )2

1dx

x 1 x 1− +∫

Solution :

(a) ( ) ( )2 2

3 2x x 1 x x 1

x x 3 x 2x x 6x− − − −=

− +− −

Let ( ) ( )2x x 1 A B C

x x 3 x 2 x x 3 x 2− − = + +

− + − + (1)

Multiplying both sides by ( ) ( )x x 3 x 2− + , we have

( ) ( ) ( ) ( )2x x 1 A x 3 x 2 Bx x 2 Cx x 3− − = − + + + + −

Putting x = 3, we get 15B = 5 or1

B3

=

Putting x 0= , we get 6A 1− = − or1

A6

=

Putting x 2= − , we get 10C 5= or1

C2

=

Substituting these values in (1), we have

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

x x 3 x 2 6x 3 x 3 2 x 2− − = + +

− + + +

MATHEMATICS 393

Notes

MODULE - VCalculus

Integration

⇒ ( ) ( )2

3 2x x 1 1 1 1

dx dx dx dx6x 3 x 3 2 x 2x x 6x

− − = + +− +− −∫ ∫ ∫ ∫

1 1 1

log x log x 3 log x 2 C6 3 2

= + − + + +

(b) ( ) ( ) ( ) ( )( ) ( ) ( )221 1 1

x 1 x 1 x 1x 1 x 1 x 1 x 1= =

+ − +− + − +

Let ( )( ) ( )2 21 A B C

x 1 x 1x 1 x 1 x 1= + +

− +− + +

Multiplying both sides by ( ) ( )2x 1 x 1− + , we have

( ) ( )( ) ( )21 A x 1 B x 1 x 1 C x 1= + + − + + −

Putting x = 1, we get1

A4

=

Putting x 1= − we get,1

C2

= −

0 A B= +

⇒1

B4

= −

∴ ( ) ( ) ( ) ( )22

1 1 1 1 1 1dx dx dx dx

4 x 1 4 x 1 2x 1 x 1 x 1= − −

− +− + +∫ ∫ ∫ ∫

1 1 1 1log x 1 log x 1 C

4 4 2 x 1 = − − + − − + +

( )1 1 1

log x 1 log x 1 C4 4 2 x 1

= − − + + ++

Example 26.49 Evaluate :3

2x x 1

dxx 1

+ +−∫

Solution : Let 3

2x x 1

I dxx 1

+ +=

−∫

Now ( ) ( )3

2 2x x 1 2x 1 2x 1

x xx 1 x 1x 1 x 1

+ + + += + = ++ −− −

∴ ( )( )2x 1

I x dxx 1 x 1

+= + + −

∫ (1)

MATHEMATICS

Notes

MODULE - VCalculus

394

Integration

Let ( ) ( )2x 1 A B

x 1 x 1 x 1 x 1+ = +

+ − + − (2)

⇒ ( ) ( )2x 1 A x 1 B x 1+ = − + +

Putting x = 1, we get 3

B2

=

Putting x 1= − , we get1

A2

=

Substituting the values of A and B in (2) and integrating, we have

( ) ( )22x 1 1 1 3 1

dx dx dx2 x 1 2 x 1x 1

+ = ++ −−∫ ∫ ∫

1 3

log x 1 log x 12 2

= + + − (3)

∴ From (1) and (3), we have2x 1 3

I log x 1 log x 1 C2 2 2

= + + + − +


(a)( ) ( )2

8dx

x 2 x 4+ +∫ (b)( )3

1dx

x x 1+∫

Solution :

(a) ( ) ( ) 228 A Bx C

x 2 x 4x 2 x 4

+= ++ ++ +

(As 2x 4+ is not factorisable into linear factors)

Multiplying both sides by ( ) ( )2x 2 x 4+ + , we have

( ) ( ) ( )28 A x 4 Bx C x 2= + + + +

On comparing the corresponding coefficients of powers of x on both sides, we get

0 A B0 2B C A 1, B 1, C 28 4A 2C

= + = + ⇒ = = − == +

∴ ( ) ( ) 228 1 x 2

dx dx dxx 2 x 4x 2 x 4

−= −+ −+ +∫ ∫ ∫

2 41 1 2x dx

dx dx 2x 2 2 x 4 x 4

= − ++ + +∫ ∫ ∫

MATHEMATICS 395

Notes

MODULE - VCalculus

Integration

2 11 1 xlog x 2 log x 4 2 tan C

2 2 2−= + − + + ⋅ +

2 11 xlog x 2 log x 4 tan C

2 2−= + − + + +

(b) ( ) ( ) ( ) 23 21 1 A B Cx D

x x 1 x x 1x x 1 x x 1 x x 1

+= = + ++ − ++ + − +

Multiplying both sides by ( )3x x 1+ or ( )( )2x x 1 x x 1+ − + , we have

( )( ) ( ) ( ) ( ) ( )2 21 A x 1 x x 1 Bx x x 1 Cx D x x 1= + − + + − + + + +

On comparing the corresponding coefficients of powers of x on both sides, we get

0 A B C0 B C D 1 2 1

A 1, B , C , D0 B D 3 3 31 A

= + + = − + + ⇒ = = − = − == + =

∴ ( ) ( ) ( )3 2

1 1 1 2x 1x 3 x 1x x 1 3 x x 1

−= − −++ − +

∴( ) 23

1 1 1 1 1 2x 1dx dx dx dx

x 3 x 1 3 x x 1x x 1

−= − −+ − ++∫ ∫ ∫ ∫

21 1log x log x 1 log x x 1 C

3 3= − + − − + +


dxsin x sin2x−∫

Solution :

Let dx dx

Isinx sin2x sinx 2s inxcosx

= =− −∫ ∫

( )dx

sinx 1 2 c o s x=

−∫Multiplying numerator and denominator by sin x, we have

( ) ( ) ( )2 2sin x d x sin x dx

Isin x 1 2cosx 1 cos x 1 2cosx

= =− − −∫ ∫

( ) ( )( )sinx dx

1 cosx 1 cosx 1 2 c o s x=

+ − −∫

MATHEMATICS

Notes

MODULE - VCalculus

396

Integration

Put cosx t= , then sinx dx dt− =

∴ ( ) ( ) ( )dt

I1 t 1 t 1 2t

−=+ − −∫

Let ( ) ( ) ( )1 A B C

1 t 1 t 1 2t 1 t 1 t 1 2t− = + +

+ − − − + −

Then, ( ) ( ) ( ) ( ) ( ) ( )1 A 1 t 1 2t B 1 t 1 2t C 1 t 1 t− = + − + − − + − +

Putting t 1= − , we get 1B

6= −

Putting t 1= , we get1

A2

=

Putting1

t2

= , we get4

C3

= −

∴( )

1 1 1 1 4 dtI dt dt

2 1 t 6 1 t 3 1 2t= − −

− + −∫ ∫ ∫

1 1 4

log 1 t log 1 t log 1 2t C2 6 6

= − − − + + − +

1 1 2

log 1 cosx log 1 cosx log 1 2cosx C2 6 3

= − − − + + − +


22sin2 cos

d4 cos 4sin

θ − θθ

− θ − θ∫Solution :

Let 22sin2 cos

I d4 cos 4sin

θ − θ= θ

− θ − θ∫

( )

24sin 1 cos d

3 sin 4sin

θ − θ θ=

+ θ − θ∫Let sin t,θ = then cos d dtθ θ =

∴2

4t 1I dt

3 t 4t

−=

+ −∫

Let 24t 1 A B

t 3 t 13 t 4t

−= +

− −− −Thus ( ) ( )4t 1 A t 1 B t 3− = − + −

Put t 1= then 3B

2= −

MATHEMATICS 397

Notes

MODULE - VCalculus

Integration

Put t 3= then 11

A2

=

∴ 11 1 3 dt

I dt2 t 3 2 t 1

= − − − ∫ ∫

11 3

log t 3 log t 1 C2 2

= − − − +

11 3

log sin 3 log sin 1 C2 2

= θ − − θ − +

Example 26.53 Evaluate :3

3tan tan

1 tan

θ + θ+ θ∫

Solution :

Let ( )23

3 3

tan 1 tantan tanI d d

1 tan 1 tan

θ + θθ + θ= θ = θ+ θ + θ∫ ∫

2

3tan sec

d1 tan

θ θ= θ

+ θ∫Let tan tθ = , then 2sec d dtθ θ =

∴ ( ) ( )3 2

t dt t dtI

1 t 1 t 1 t t= =

+ + − +∫ ∫ (1)

Let ( ) ( ) 22t A Bt C

1 t 1 t t1 t 1 t t

+= ++ − ++ − +

Then ( ) ( ) ( )2t A 1 t t Bt C 1 t= − + + + +

Comparing the coefficients of t, we getA B 0,+ = A B C 1,− + + = A C 0+ =

⇒1

A3

= − , 1

B3

= , 1

C3

=

∴ 21 1 1 t 1

I dt dt3 1 t 3 1 t t

+= − +

+ − +∫ ∫

21 dt 1 2t 2

dt3 1 t 6 t t 1

+= − +

+ − +∫ ∫

( )2

2t 1 31 dt 1dt

3 1 t 6 t t 1

− += − +

+ − +∫ ∫

MATHEMATICS

Notes

MODULE - VCalculus

398

Integration

( )2 22

2t 1 dt1 dt 1 1 13 1 t 6 2t t 1 1 3

t2 2

−= − + +

+ − + − +

∫ ∫ ∫

2 1

1t1 1 1 2 2log 1 t log t t 1 tan

3 6 2 3 32

−

−

= − + + − + + ⋅

2 11 1 1 2t 1

log 1 t log t t 1 tan C3 6 3 3

− − = − + + − + + +

21 1log 1 tan log tan tan 1

3 6= − + θ + θ − θ + 11 2tan 1

tan C3 3

− θ − + +


Evaluate the following :

1. (a) 24x 5 dx−∫ (b) 2x 3x dx+∫ (c) 23 2x 2x dx− −

2. (a)( ) ( )

x 1dx

x 2 x 3+

− −∫ (b) 2

xdx

x 16−∫

3. (a)3

2x

dxx 4−∫ (b)

( ) ( )

2

22x x 1

dxx 1 x 2

+ +

− +∫

4. ( )

2

3x x 1

dxx 1

+ +

−∫

5. (a) s i n xdx

sin 4 x∫ (b)( )

1 cosxdx

cosx 1 cosx−

+∫

LET US SUM UP

• Integration is the inverse of differentiation

• Standard form of some indefinite integrals

(a) nx dx∫ n 1xC

n 1

+= ++

( )n 1≠ −

LET US SUM UP

MATHEMATICS 399

Notes

MODULE - VCalculus

Integration

(b) 1

dxx∫ l o g | x | C= +

(c) sinxax∫ cosx C=− +

(d) cosxdx∫ sinx C= +

(e) 2sec xdx∫ tanx C= +

(f) 2cosec xdx∫ cotx C= − +

(g) secxtanxdx∫ secx C= +

(h) cosecxcotxdx∫ cosecx C= − +

(i) 2

1dx

1 x−∫ 1sin x C−= +

(j) 2

1dx

1 x+∫ 1tan x C−= +

(k) 2

1dx

x x 1−∫ 1sec x C−= +

(l) xe dx∫ xe C= +

(m) xa dx∫ ( )xa

C a 0anda 1loga

= + > ≠

• Properties of indefinite integrals

(a) ( ) ( )f x g x dx± ∫ ( ) ( )f x dx g x dx= ±∫ ∫(b) ( )kf x dx∫ ( )k f x dx= ∫

(i) ( )nax b dx+∫( ) ( )

n 1ax b1C n 1

a n 1

++= + ≠ −

+

(ii) 1

dxax b+∫

1log ax b C

a= + +

(iii) ( )sin ax b dx+∫ ( )1cos ax b C

a−= + +

(iv) ( )cos ax b dx+∫ ( )1sin ax b C

a= + +

(v) ( )2sec ax b dx+∫ ( )1tan ax b C

a= + +

MATHEMATICS

Notes

MODULE - VCalculus

400

Integration

(vi) ( )2cosec ax b dx+∫ ( )1cot ax b C

a= − + +

(vii) ( ) ( )sec ax b tan ax b dx+ +∫ ( )1sec ax b C

a= + +

(viii) ( ) ( )cosec ax b cot ax b dx+ +∫ ( )1cosec ax b C

a= − + +

(ix) ax be dx+∫ ax b1e C

a+= +

( )( )

f ` xdx

f x∫ ( )log f x C= +

(i) tanxdx∫ log cosx C= − + logsecx C= +

(ii) cotxdx∫ log sinx C= +

(iii) secxdx∫ log secx tanx C= + +

(iv) cosecxdx∫ log cosecx cotx C= − +

• (i) 2 21 1 a x

dx log C2a a xa x

+= +

−−∫ (ii) 2 21 1 x a

dx log C2a x ax a

−= +

+−∫

(iii) 1

2 21 1 x

dx tan Ca ax a

−= ++∫ (iv) 1

2 2

1 xdx sin C

aa x

−= +−

∫

(v) 2 22 2

1dx log x x a C

x a= + − +

−∫

(vi) 2 22 2

dxlog x x a C

x a= + + +

+∫

• (i)2

14

1xx 1 1 xdx tan C

2 2x 1−

− += +

+

∫

(ii)2

4

1x 2x 1 1 xdx log C

12 2x 1 x 2x

+ −−= +

+ + +∫

(iii)2

14

1 1x x 2x 1 2 xdx tan log C12 2 2x 1 x 2x

− − + −

= + + + + +

∫

MATHEMATICS 401

Notes

MODULE - VCalculus

Integration

(iv) 14

1 1x x 21 1 12 xdx tan log C

12 2 2 2 2x 1 x 2x

− − + −

= − + + + +

∫

• (i)2

14 2

1xx 1 1 xdx tan C

3 3x x 1−

− += +

+ +

∫

(ii)2

4 2

1x 1x 1 1 xdx log C

12x x 1 x 1x

+ −−= +

+ + + +∫

(iii) 1 14 2

1 1 1 tanx 1 1 cotx 1dx tan tan C

2 2 2 2 2x x 1− − − − = + + + +

∫

(iv) ( ) ( )1tanx cotx dx 2sin sinx cosx C−+ = − +∫• Integral of the product of two functions

I function ×Integral of II function — [ ]DerivativeofIfungtion Integralof IIfungtion dx×∫• ( ) ( ) ( )x xe f x f ` x dx e f x C+ = + ∫

•2 2 2 2 2 11 x

a x dx x a x a sin C2 a

− − = − + + ∫

2 2 22 2 2 2x x a a

x a dx log x x a C2 2

−− = − + − +∫

2 2 22 2 2 2x a x a

a x dx log x a x C2 2+

+ = + + + +∫• Rational fractions are of following two types :(i) Proper, where degree of variable of numerator < denominator.(ii) Improper, where degree of variable of numerator ≥ denominator..

. • If ( )g x is a proper fraction ( )( )

f xg x can be resolved into real factors, then

( )( )

f xg x can be

written in the following form :

MATHEMATICS

Notes

MODULE - VCalculus

402

Integration

Factors in denominator Corresponding partial fraction

ax b+A

ax b+

( )2ax b+ ( )2A B

ax b ax b+

+ +

( )3ax b+ ( ) ( )2 3A B C

ax b ax b ax b+ +

+ + +

2ax bx c+ + 2Ax B

ax bx c

++ +

( )22ax bx c+ + ( )2 22

Ax B Cx D

ax bx c ax bx c

+ +++ + + +

where A, B, C, D are arbitrary constants.

l http://www.wikipedia.orgl http://mathworld.wolfram.com

TERMINAL EXERCISE

Integrate the following functions w.r.t. x :

1.3 3

2 2sin x cos x

sin xcos x

+2. 1 sin2x+

3. 2 2cos2x

cos xsin x4. ( )2tanx cotx−

5. 2 2

4 1

1 x 1 x−

+ −6.

22sin x1 cos2x+

7.22cos x

1 cos2x−8.

2x xsin cos

2 2 +

9.2x x

cos sin2 2

− 10. ( )cos 7x − π

11. ( )sin 3x 4+ 12. ( )2sec 2x b+

SUPPORTIVE WEB SITES

MATHEMATICS 403

Notes

MODULE - VCalculus

Integration

13. dxsinx cosx−∫ 14. ( )2 1

1dx

1 x tan x−+∫

15.cosecx

dxxlog tan2

∫ 16.co tx

dx3 4logsinx+∫

17. dxsin2xlogtanx∫ 18.

x

xe 1

dxe 1

+−∫

19. 4sec xtanxdx∫ 20. x xe sine dx∫

21.2

xdx

2x 3+∫ 22.

2sec xdx

tanx∫

23. 225 9x dx−∫ 24. 22ax x dx−∫25. 23x 4dx+∫ 26. 21 9x dx+∫

27.2

2 2

x dx

x a−∫ 28. 2 2

dx

sin x 4cos x+∫

29.dx

2 cosx+∫ 30. 2dx

x 6x 13− +∫

31. 2dx

1 3sin x+∫ 32.2

2 2x

dxx a−∫

33. 4

dx

x 9 x+∫ 34.

s inxdx

sin3x∫

35. 2dx

1 4cos x−∫ 36. ( )2sec ax b dx+∫

37. ( )dx

x 2 logx+∫ 38.5

6x

dx1 x+∫

39.cosx sinx

dxsinx cosx

−+∫ 40.

cotxdx

logsinx∫

41.2sec x

dxa btanx+∫ 42.

s inxdx

1 cosx+∫43. 2cos xdx∫ 44. 3sin xdx∫45. sin5xsin3x dx∫ 46. 2 3sin x cos x dx∫47. 4sin xdx∫ 48.

1dx

1 s i n x+∫

MATHEMATICS

Notes

MODULE - VCalculus

404

Integration

49. 3tan x dx∫ 50.cosx s i n x

dx1 sin2x

−+∫

51.2cosec x

dx1 cot x+∫ 52. 2

1 x cos2xdx

x sin2x 2x

+ ++ +∫

53.sec cosec d

log tanθ θ θ

θ∫ 54.cot dlog sin

θ θθ∫

55. 2dx

1 4x+∫ 56.1 tan

d1 tan

− θ θ+ θ∫

57.1x

21

e dxx

−∫ 58. 2 2 2 2

sinxcosx dx

a sin x b cos x+∫

59.dx

sin x cosx+∫ 60.x 1

2

1e cos x dx

1 x−

− −

∫

61.x

2sinx cosx

e dxcos x

+ ∫ 62. 1 1 cosx

tan dx1 cosx

− −+∫

63.1 1 x

cos 2cot dx1 x

− − +

∫ 64.

( )

1

32 2

sin xdx

1 x

−

−∫

65. x logx dx∫ 66. ( ) ( )x xe 1 x log xe dx+∫

67. ( )2log x

dx1 x+∫ 68. x 2e sin x dx∫

69. ( )cos log x dx∫ 70. ( )log x 1 dx+∫71.

( ) ( )

2

2x 1

dxx 1 x 3

+

− +∫ 72. 2sin cos

dcos cos 2

θ θθ

θ − θ −∫

73. ( )5dx

x x 1+∫ 74. ( )( )2

2 2x 1

dxx 2 2x 1

++ +∫

75. ( )( )logx

dxx 1 log x 2 log x+ +∫ 76. x

dx

1 e−∫

MATHEMATICS 405

Notes

MODULE - VCalculus

Integration

ANSWERS


1.7 7 7 7 72 2 2 2 22 2 2 2 2

x 1, x 2, x 3, x 4, x 57 7 7 7 7

+ + + + +

2. (a)6x

C6

+ (b) sinx C+ (c) 0

3. (a)7x

C7

+ (b) 61

C6x

+ (c) log x C+

(d)

x35 C

3log

5

+

(e)433

x C4

+ (f) 81

C8x

−+

(g) 2 x C+ (h)199x C+

4. (a) cosec + C− θ (b) sec Cθ +(c) tan Cθ + (d) cot C− θ +


1. (a)2x 1

x C2 2

+ + (b) 1x tan x C−− + +

(c)3

10 22x x 2 x C

3− + + (d) 5 4 3

1 3 2 78x C

xx 4x 3x− − + + − +

(e)3

1xx tan x C

3−− − + (f)

2x4x 4logx C

2+ + +

2. (a)1

tanx C2

+ (b) tan x x C− +

(c) 2 cosec x C− + (d)1

cotx C2

− +

(e) sec x C− + (f) cot x cosec x+C− +

3. (a) 2 sinx C+ (b) 2 cosx C− +

(c)1

cotx C2

− +

4. (a) ( )32

2x 2 C

3+ + (b) ( )16

17 1C

16 x 1

− ++

MATHEMATICS

Notes

MODULE - VCalculus

406

Integration


1. (a) ( )1cos 4 5x C

5− + (b) ( )1

tan 2 3x C3

+ +

(c) log sec x tan x C4 4π π + + + +

(d) ( )1sin 4x 5 C

4+ +

(e) ( )1sec 3x 5 C

3+ + (f) ( )1

cosec 3 5x C5

− + +

2. (a) ( )31

C12 3 4x

+− (b) ( )51

x 1 C5

+ +

(c) ( )1114 7x C

77− − + (d) ( )41

4x 5 C16

− +

(e)1

log 3x 5 C3

− + (f)2

5 9x C9

− − +

(g) ( )312x 1 C

6+ + (h) log x 1 C+ +

3. (a) 2x 11e C

2+ + (b) 3 8x1

e C8

−− +

(c) ( )7 4x1

C4e +− +

4. (a)1 sin2x

x C2 2

+ + (b)1 3 1

cos2x cos6x C32 2 6

− + +

(c)1 cos7x

cosx C2 7

− − + (d)1 sin6x sin 2 x

C2 6 2

+ +


1. (a) 21log 3x 2 C

6− + (b) 2log x x 1 C+ + +

(c) 2log x 9x 30 C+ + + (d) 31log x 3x 3 C

3+ + +

(e) 2log x x 5 C+ − + (f) 2log 5 x C+ +

(g) log 8 log x C+ +

2. (a) x1log a be C

b+ + (b) ( )1 xtan e C− +

MATHEMATICS 407

Notes

MODULE - VCalculus

Integration


1. (a) 3 x 3x log C

2 x 3−

+ ++

(b) ( )1 xtan e C− +

(c) ( )1 21tan x C

2− + (d)

11 3xsin C

3 4− +

(e) ( )11tan 2tanx C

2− + (f) 1 x 1

sin C2

− + +

(g)11 x 1

tan c3 6 6

− + + (h) 1 x 2sin C

3− + +

(i)11 x

sec C22 3

− + (j)2

11 tan 1tan C

2 2 tan− θ −

+ θ

(k) x 2xlog e 1 e C+ + + (l) 1 2sin x 1 x C− − − +

(m)1 x a

sin Ca

− − + (n)1 31 4

sin x C4 3

− +

(o) 2 2x 1 log x x 1 C+ + + + + (p) 21 2x 9 4x

log C2 2

+ ++

(q) 21log 2cos 4cos 1 C

2− θ + θ − +

(r) 2log tan x tan x 4 C+ − +

(s) 1 x 2tan C

1− + +

(t)2

21 5log x x C

4 4 + + +

CHECK YOUR PROGRESS 26.61. (a) xcosx sinx C− + +

(b) ( )21 xcos2x sin2x1 x sin2x C

2 2 4+ + − +

(c) xcos2x 1 sin2x

C2 2 2

− + +

2. (a) xtanx log secx x C− − +

(b) 3 21 1 1 1x x sin2x xcos2x sin2x C

6 4 4 8− − + +

MATHEMATICS

Notes

MODULE - VCalculus

408

Integration

3. (a)4 4x log2x x

C4 16

− + (b)3 3x x

x logx x C3 9

− − + +

(c) ( )2x logx 2xlogx 2x C− + +

4. (a) ( )

1 n 1 n

2x x

logx C1 n 1 n

− −− +

− − (b) ( )logx. log logx 1 C− +

5. (a)2

3x x 2x 2e C

3 9 27

− + +

(b)

4x 4xe ex C

4 16− +

6. ( )2

2x 1logx logx C

2 2 − + +

7. (a)1 2xsec x log x x 1 C− − + − +

(b)2

1 1x x 1cot x cot x C

2 2 2− −+ + +


1. (a) xe secx C+ (b) xe logsecx tanx C+ +

2. (a) x1e C

x+ (b) x 1e sin x C− +

3. ( )

x

2e

C1 x

++ 4.

xeC

1 x+

+

5.x

xtan C2

+

6. ( )x1e sin2x 2cos2x C

5− +


1. (a)2 25 5 5

x x log x x C4 4 4

− − + − +

(b) ( ) 2 22x 3 9 3x 3x log x x 3x C

4 8 2+ + − + + + +

(c) ( ) 2 11 7 2x 12x 1 3 2x 2x sin C

4 4 2 7− + + − − + +

MATHEMATICS 409

Notes

MODULE - VCalculus

Integration

2. (a) 4log |x 3 | 3log |x 2 | C− − − +

(b)1

log |x 4 | log |x 4 | C2

− + + +

3. (a) [ ]2x

2 log|x 2 | log|x 2 | C2

− − + + +

(b) ( )( )

11 7 4log|x 1 | log x 2 C

9 9 3 x 1− + + − +

−

4. ( ) ( )23 3

log|x 1| Cx 1 2 x 1

− − − +− −

5. (a)1 1

log1 sinx 1 sinx8 8

− − +

1 1log 1 2sinx log1 2sinx C

4 2 4 2− − + + +

(b)x

log secx tanx 2tan C2

+ − +

TERMINAL EXERCISE1. secx cosecx C− + 2. sinx cosx C− +3. cotx tanx C− − + 4. tanx cotx 4x C− − +

5. 1 14tan x sin x C− −− + 6. tanx x C− +

7. cotx x C− − + 8. x cosx C− +

9. x cosx C+ + 10.( )sin 7x

C7

− π+

11.( )cos 3x 4

C3

− ++ 12.

( )tan 2x bC

2+

+

13.1

log cosec x cot x C4 42π π − − − +

14. 1log tan x C− + 15.x

log logtan C2

+

16.1

log 3 4logsinx C4

+ + 17.1

log logtanx C2

+

18.x x2 22log e e C

−

− + 19. 41sec x C

4+

MATHEMATICS

Notes

MODULE - VCalculus

410

Integration

20. xcose C− + 21.22x 3

C2

+ +

22. 2 tanx C+

23. ( )2 11 25 3x 25 9x sin x C

6 6 5− − + +

24. ( ) 2 2 11 1 x ax a 2ax x a sin C

2 2 a− − − − + +

25.2 2x 3x 4 2 3x x 4

log C2 23

+ + ++ +

26.2

2x 9x 1 1log 3x 1 9x C

2 6+

+ + + +

27.2 2 2 2 21 1

x x a a log x x a C2 2

− + + − +

28.11 tanx

tan C2 2

− + 29.

1

xtan

2 2tan C

3 3−

+

30.11 x 3

tan C2 2

− − + 31. ( )11

tan 2tanx C2

− +

32.a x a

x log C2 x a

−+ +

+ 33.4

4

1 9 x 3log C

12 9 x 3

+ −+

+ +

34.1 3 tanx

log C2 3 3 tanx

++

− 35.1 tanx 2

log C2 2 tanx 2

−+

+

36. ( )1tan ax b C

a+ + 37. ( )log 2 logx C+ +

38. ( )61log 1 x C

6+ + 39. log sinx cosx C+ +

40. ( )loglog sinx C+ 41.1

log a btanx Cb

+ +

42. log 1 cosx C− + + 43.1 sin2x 1

x C2 2 2

+ +

MATHEMATICS 411

Notes

MODULE - VCalculus

Integration

44.3cos x

cosx C3

− + + 45.1 sin2x 1 sin8x

C2 2 2 8

− +

46.5

31 sin xsin x C

3 5− + 47. [ ]1

12x 8sin2x sin4x C32

− + +

48. tanx secx C− + 49.2tan x

log cosx C2

+ +

50.1

Ccosx sinx

− ++

51.1

log C1 cotx

++

52. 21log x sin2x 2x C

2+ + + 53. log tan Cθ +

54. loglogsin Cθ + 55. 11tan 2x

2−

56. log cos sin Cθ+ θ + 57.1xe C

−+

58. ( )2 2 2 2

2 2

1log a sin x b cos x C

2 a b+ +

−

59.1

log sec x tan x C4 42π π − + − +

60. x 1e cos x C− + 61. xe secx C+

62. 21x C

4+ 63. 21

x C2

− +

64.1

22

xsin x 1log 1 x C

21 x

−+ − +

−65.

322 2

x logx C3 3

− +

66. ( )x xxe log xe 1 C − +

67.1

log x log x log x 1 C1 x

− + − + ++

68. ( )x

x1 ee 2sin2x cos2x C

2 10− + +

69. ( ) ( )xcos logx sin logx C

2+ +

70. xlog x 1 x log x 1 C+ − + + +

71. ( )3 1 5

log x 1 log x 3 C8 2 x 1 8

− − + + +−

MATHEMATICS

Notes

MODULE - VCalculus

412

Integration

72.2 1

log cos 2 log cos 1 C3 3

− θ− − θ+ +

73.5

51 x

log C5 x 1

++

74. ( )1 11 xtan tan 2x C

3 2 2− − + +

75.( )22 logx

log C1 logx+

++

76.x

xe

log C1 e

+−

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