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Topic : Idefinite & Definite Integration Available Online :

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1.

2.

(i) (iii) (v) (vii) (ix) (xi) (xii) (xiii) (xiv) (xv) (xvii) (xix)

If f & g are functions of x such that g(x) = f(x) then,

f(x) dx = g(x) + c d {g(x)+c} = f(x), where c is called the constant of integration. dx

Standard Formula:

ax bn1

(ax + b)n dx = a n 1 + c, n 1

(ii)

dx 1 = ln (ax + b) + c ax b a

1 eax+b dx = eax+b + c a

(iv)

apx+q

dx

=

1 p

apxq n a

+ c; a > 0

sin (ax + b) dx = 1 cos (ax + b) + c a

(vi)

1 cos (ax + b) dx = sin (ax + b) + c

a

tan(ax + b) dx = 1 ln sec (ax + b) + c a

(viii)

cot(ax + b) dx = 1 ln sin(ax + b)+ c

a

1 sec? (ax + b) dx = tan(ax + b) + c a

(x)

1 sec (ax + b). tan (ax + b) dx = sec (ax + b) + c a

cosec?(ax + b) dx = 1 cot(ax + b)+ c a

cosec (ax + b). cot (ax + b) dx = 1 cosec (ax + b) + c a

secx dx = ln (secx + tanx) + c

OR

ln

tan

4

2x

+

c

cosec x dx = ln (cosecx cotx) + c OR ln tan x + c OR ln (cosecx + cotx) + c 2

d x = sin1 x + c

a2 x2

a

(xvi)

dx

1

=

tan1 x

+ c

a2 x2 a

a

dx

1

=

sec1 x

+ c

x x2 a2 a

a

(xviii)

d x = ln x x2 a2 x2 a2

OR sinh1 x + c a

d x = ln x x2 a2 x2 a2

OR

cosh1 x + c

a

(xx)

d x 1 ax

a2 x2

=

2a

ln

ax

+ c

(xxi)

dx

1

xa

x2 a2

=

2a

ln

xa

+ c

(xxii)

a2 x2

dx

x

=

a2 x2

a2

+

sin1 x + c

2

2

a

(xxiii)

x2 a2 dx = x 2

x2 a2

a2

+

2

n

x

x2 a2

a

+ c

(xxiv)

x2 a2 dx = x 2

x2 a2

a2

2

n

x

x2 a2

a

+c

(xxv)

e ax eax. sin bx dx = a2 b2

(a sin bx b cos bx) + c

(xxvi)

eax eax. cos bx dx = a2 b2 (a cos bx + b sin bx) + c

3. Theorems on integration

(i)

(iii) Note : (i)

(ii)

c f(x).dx = c f(x).dx

(ii)

(f(x) g(x))dx = f(x)dx g(x)dx

f(x)dx g(x) c

f(ax b)dx

=

g(ax b) a

+ c

every contineous function is integrable

the integral of a function reffered only by a constant.

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f(x).dx = g(x) + c = h(x) + c

g(x) = f(x) & g(x) ? h(x) = 0 means, g(x) ? h(x) = c

h(x) = f(x)

Example : Solution.

Evaluate : 4x5 dx

4x5 dx

=

4 6

x6 + C =

2 3

x6 + C.

Example : Solution.

Example : Solution.

Example : Solution. Example: Solution.

Evaluate :

x3

5x2

4

7 x

2 x

dx

x3

5x2

4

7 x

2 x

dx

= x3 dx + 5x2 dx ? 4dx + 7 dx + x

2 dx x

= x3 dx + 5 . x2 dx ? 4 . 1. dx + 7 . 1 dx + 2 . x1/ 2dx x

=

x4 4

+ 5 .

x3 3

?

4x

+

7

log

|

x

|

+

2

x1/ 2 1/ 2

+

C

=

x4 4

+

5 3 x3

? 4x + 7 log | x | + 4

x +C

Evaluate : ex loga ea log x ealoga dx

We have,

ex loga ea log x ealoga dx

= elogax elog xa elogaa dx

= (ax xa aa ) dx

= ax dx + xa dx + aa dx

ax

x a 1

= loga + a 1 + aa . x + C.

2x 3x

Evaluate :

5x dx

2x 3x 5x dx

=

2x 5x

3x 5x

dx

=

2

x

3

x

5 5

(2 / 5)x

(3 / 5)x

dx = loge 2 / 5 + loge 3 / 5 + C

Evaluate : sin3 x cos3 x dx

1

= 8

(2sin x cos x)3 dx

1

= 8

sin3 2x dx

1

= 32

(3 sin2x sin 6x) dx

1 3 sin2x sin6x

= 8

4

dx

1 = 32

3 2

cos

2x

1 6

cos

6x

+

C

Example :

x4

Evaluate : x2 1 dx

Solution.

x4 x2 1 dx

=

x4 11 x2 1 dx =

x4 1

1

x2 1 + x2 1 dx =

(x2 1) dx +

1

x3

x2 1 dx = 3 ? x + tan?1 x + C

Example: Solution.

1

Evaluate : 4 9x2 dx

We have

1 4 9x2

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1

1

= 9 4 x2 dx

9

1

1

= 9 (2 / 3)2 x2 dx

=

1 9

.

1 (2 / 3)

tan?1

x 2/3

+ C =

1 6

tan?1

3x 2

+ C

Example :

cos x cos 2x dx

Solution.

cos x cos2x dx

1

=

2cos x cos 2x dx

2

1

=

(cos3x cos x) dx

2

Self Practice Problems

=

1 2

sin3x sin x

3

1

+ c

1.

Evaluate : tan2 x dx

Ans. tanx ? x + C

2.

1

Evaluate : 1 sin x dx

Ans. tanx ? sec x + C

4. Integration by Subsitutions

If we subsitute x = (t) in a integral then

(i) everywhere x will be replaced in terms of t. (ii) dx also gets converted in terms of dt.

(iii) (t) should be able to take all possible value that x can take.

Example : Solution.

Evaluate : x3 sin x4 dx

We have

= x3 sin x4 dx

Example : Solution.

Let x4 = t

(n x)2 dx x

(n x)2 dx x

Put nx = t

=

t2. dx x

t3 = +c

3

d(x4) = dt

1

4x3 dx = dt dx = 4x3 dt

1

dx = dt

x

= t2dt

(n x)3

=

+ c

3

Example : Solution.

Example : Solution.

Evaluate (1 sin2 x)cos x dx

Put sinx = t cosx dx = dt

(1 t2 ) dt = t + t3 + c 3

sin3 x

= sin x +

+ c

3

x

Evaluate : x4 x2 1 dx We have,

x

x

= x 4 x2 1 dx = (x2 )2 x2 1 dx

dt

Let x2 = t, then, d (x2) = dt

2x dx = dt

dx =

2x

x

dt

= t2 t 1 . 2x

1

1

= 2

t2 t 1 dt

=

1 2

1

t

1 2

2

3 2 2

dt

1 = 2.

1 3

tan?1

t 1

2 3

+

C

2

2

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=

1 3

tan?1

2 t

3

1

+

C =

1 3

tan?1

2x2 1 3

+ C.

Note: (i) (iii) (iv)

[ f(x)]n f (x) dx = (f(x))n1 n1

(ii)

f (x)

f (x)n

d x n N Take xn common & put 1 + xn = t.

x (xn 1)

dx

(n1)

n N, take xn common & put 1+xn = tn

x2 xn 1 n

( f ( x))1n dx =

1 n

(v)

xn

dx 1 xn 1/ n

take xn common as x and put 1 + xn = t.

Self Practice Problems

1.

sec 2 x dx 1 tan x

Ans. n |1 + tan x| + C

2.

sin(nx) dx x

Ans. ? cos (n x) + C

5. Integration by Part :

f(x) g(x) dx = f(x)

g(x) dx ?

d f(x) g(x) dx dx

dx

(i)

when you find integral g(x)dx then it will not contain arbitarary constant.

(ii)

g(x)dx should be taken as same both terms.

(iii) the choice of f(x) and g(x) is decided by ILATE rule.

the function will come later is taken an integral function.

Inverse function

L

Logrithimic function

A

Algeberic function

T

Trigonometric function

E

Exponential function

Example : Solution.

Example : Solution.

Evaluate : x tan1 x dx

x tan1 x dx

= (tan?1 x) x2 ? 2

1

x2

1 x2 . 2 dx

x2

1

=

tan?1 x ?

2

2

x2 11

x2

1

x2 1

dx = 2

tan?1 x ? 2

1

1 x2 1

dx

x2

1

=

tan?1 x ? [x ? tan?1 x] + C.

2

2

Evaluate : x log(1 x) dx

x log(1 x) dx

x2

1 x2

= log (x + 1) . ?

. dx

2

x1 2

x2

1

= log (x + 1) ?

x2

x2

1

dx = log (x + 1) ?

x2 1 1 dx

2

2 x1

2

2 x 1

x2

1 x2 1

1

= log (x + 1) ?

+

dx

2

2 x 1 x1

=

x2 2

log (x + 1) ?

1 2

( x

1)

x

1

1

dx

=

x2 2

log (x + 1) ?

1 2

x2

2

x log | x 1|

+ C

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Example :

Evaluate : e2x sin3x dx

Solution.

Let = e2x sin3x dx. Then,

= e2x sin3x dx

= e2x

cos3x 3

?

2e2x

cos3x 3

dx

1

2

= ? 3 e2x cos 3x + 3

e2x cos 3x dx

1

2

= ? 3 e2x cos 3x + 3

e2x

sin 3x 3

2e2x

sin 3x 3

dx

1

2

4

= ? 3 e2x cos 3x + 9 e2x sin 3x ? 9

e2x sin3x dx

1

2

4

= ? 3 e2x cos 3x + 9 e2x sin 3x ? 9

4

e2x

+ 9 = 9 (2 sin 3x ? 3 cos 3x)

13

e2x

e2x

9 = 9 (2 sin 3x ? 3 cos 3x)

= 13 (2 sin 3x ? 3 cos 3x) + C

Note : (i) ex [f(x) + f (x)] dx = ex. f(x) + c (ii)

Example :

ex

x (x 1)2 dx

Solution.

ex

x 11 (x 1)2 dx

Example :

ex 1 sin x dx 1 cos x

Solution.

1 2sin x cos x

ex

2 2 sin2 x

2

2

dx

ex 1 cosec2 cot x dx

2

2

Example : Solution.

n

(nx )

1 (nx)2

dx

put x = et

et

nt

1 t2

dt

x

n

(nx)

1 nx

+ c

Self Practice Problems

[f(x) + xf (x)] dx = x f(x) + c

ex

(x

1

1)

(x

1 1)2

dx

ex = (x 1) + c

x = ? ex cot + c

2

et

nt

1 1 tt

1 t2

dt

= et nt 1 + c

t

1.

x sin x dx

Ans. ? x cosx + sin x + C

2.

x2ex dx

Ans. x2 ex ? 2xex + 2ex + C

6 . Integration of Rational Algebraic Functions by using Partial Fractions:

PARTIAL FRACTIONS :

f(x) If f(x) and g(x) are two polynomials, then g(x) defines a rational algebraic function of a rational function of x.

f(x) If degree of f(x) < degree of g(x), then g(x) is called a proper rational function.

f(x) If degree of f(x) degree of g(x) then g(x) is called an improper rational function

f(x)

f(x)

If g(x) is an improper rational function, we divide f(x) by g(x) so that the rational function g(x) is expressed in the

(x) form (x) + g(x) where (x) and (x) are polynomials such that the degree of (x) is less than that of g(x).

f(x) Thus, g(x) is expressible as the sum of a polynomial and a proper rational function.

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f(x) Any proper rational function g(x) can be expressed as the sum of rational functions, each having a simple

factor of g(x). Each such fraction is called a partial fraction and the process of obtained them is called the

f(x) resolutions or decomposition of g(x) into partial fractions.

f(x) The resolution of g(x) into partial fractions depends mainly upon the nature of the factors of g(x) as

discussed below.

CASE I When denominator is expressible as the product of non-repeating linear factors.

Let g(x) = (x ? a ) (x ? a ) .....(x ? a ). Then, we assume that

1

2

n

f(x)

A1

A2

An

g(x) = x a1 + x a2 + ..... + x an

where A , A , ...... A are constants and can be determined by equating the numerator on R.H.S. to the numerat1or o2n L.H.Sn. and then substituting x = a1, a2, ........,an.

3x 2

Example : Resolve x3 6x2 11x 6 into partial fractions.

Solution.

3x 2

3x 2

We have, x3 6x2 11x 6 = (x 1)(x 2)(x 3)

3x 2

A

B

B

Let

(x 1)(x 2)(x 3) = x 1 + x 2 + x 3 . Then,

3x 2

A(x 2)(x 3) B(x 1)(x 3) C(x 1)(x 2)

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

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

3x + 2 = A(x ? 2) (x ? 3) + B (x ? 1) (x ? 3) + C(x ? 1) (x ? 2) ...........(i)

Putting x ? 1 = 0 or x = 1 in (i), we get

5 5 = A(1 ? 2) (1 ? 3) A = 2 , Putting x ? 2 = 0 or, x = 2 in (i), we obtain 8 = B (2 ? 1) (2 ? 3) B = ?8. Putting x ? 3 = 0 or, x = 3 in (i), we obtain

11 11 = C (3 ? 1) (3 ? 2) C = .

2

3x 2

3x 2

5

8

11

x3 6x2 11x 6 = (x 1)(x 2)(x 3) = 2(x 1) ? x 2 + 2(x 3)

Note : In order to determine the value of constants in the numerator of the partial fraction corresponding to the nonrepeated linear factor px + q in the denominator of a rational expression, we may proceed as follows :

q Replace x = ? p (obtained by putting px + q = 0) everywhere in the given rational expression except in the factor px + q itself. For example, in the above illustration the value of A is obtained by replacing x by 1 in

3x 2 all factors of (x 1)(x 2)(x 3) except (x ? 1) i.e.

31 2

5

A = (1 2)(1 3) = 2

Similarly, we have

321

33 2

11

B = (1 2)(2 3) = ?8 and, C = (3 1)(3 2) = 2

Example : Solution.

x3 6x2 10x 2 Resolve x2 5x 6 into partial fractions. Here the given function is an improper rational function. On dividing we get

x3 6x2 10x 2

(x 4)

x2 5x 6

= x ? 1 + (x2 5x 6)

...........(i)

x 4

x 4

we have, x2 5x 6 = (x 2)(x 3)

x 4

A

B

So, let (x 2)(x 3) = x 2 + x 3 ? x + 4 = A(x ? 3) + B(x ? 2)

Putting x ? 3 = 0 or, x = 3 in (ii), we get

1 = B(1) B = 1.

Putting x ? 2 = 0 or, x = 2 in (ii), we get

2 = A (2 ? 3) A = ? 2

...........(ii)

x 4

2

1

x3 6x2 10x 2

2

2

(x 2)(x 3) = x 2 + x 3 Hence

x2 5x 6

= x ? 1 ? x2 + x3

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CASE II When the denominator g(x) is expressible as the product of the linear factors such that some of them are repeating.

1

1

Example g(x) = (x a)k (x a1)(x a2 ).......( x ar ) this can be expressed as

A1 xa

+

A2 (x a)2

+

A3 (x a)3

+ ....+

Ak (x a)k

+

B1 (x a1)

+

B2 (x a2 )

+ ...... +

Br (x ar )

Now to determine constants we equate numerators on both sides. Some of the constants are determined

by substitution as in case I and remaining are obtained by

The following example illustrate the procedure.

Example :

3x 2

(3x 2)dx

Resolve (x 1)2(x 1)(x 2) into partial fractions, and evaluate (x 1)2(x 1)(x 2)

Solution.

Let

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

=

A1 + x 1

A2 (x 1)2

+

A3 + x 1

A4 x2

Putting

3x ? 2 = A1 (x x ? 1 = 0 or, x

? 1) (x + 1) (x +

=+1Ain3

(x (i)

? 1)2 (x we get

2) +

+ 2)

A+2A(x4

+ 1) (x + 2) (x ? 1)2 (x +

1)

.......(i)

1 1 = A2 (1 + 1) (1 + 2) A2 = 6 Putting x + 1 = 0 or, x = ?1 in (i) we get

5

?

5

=

A 3

(?2)2

(?1

+

2)

A 3

=

?

4

Putting x + 2 = 0 or, x = ?2 in (i) we get

8 ? 8 = A4 (?3)2 (?1) A4 = 9

Now equating coefficient of x3 on both sides, we get 0 = A1 + A3 + A4

5 8 13

A1 = ?A3 ? A4 = 4 ? 9 = 36

3x 2

13

1

5

8

(x 1)2(x 1)(x 2) = 36(x 1) + 6(x 1)2 ? 4(x 1) + 9(x 2)

and hence

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

13 = 36

n |x ? 1| ?

1 6(x 1)

?

5 4

n |x + 1| +

8 9

n |x + 2| + c

CASE III When some of the factors of denominator g(x) are quadratic but non-repeating. Corresponding to

Ax B each quadratic factor ax2 + bx + c, we assume partial fraction of the type ax2 bx c , where A and B are constants to be determined by comparing coefficients of similar powers of x in the numerator of both sides.

A(2ax b)

B

In practice it is advisable to assume partial fractions of the type ax2 bx c + ax2 bx c

The following example illustrates the procedure

Example :

2x 1

2x 1

Resolve (x 1)(x2 2) into partial fractions and evaluate (x 1)(x2 2) dx

Solution.

2x 1

A Bx C

Let

(x 1)(x2 2) = x 1 + x2 2 . Then,

2x 1

A(x2 2) (Bx C)(x 1)

(x 1)(x2 2) =

(x 1)(x2 2)

2x ? 1 = A (x2 + 2) + (Bx + C) (x + 1)

...(i)

Putting x + 1 = 0 or, x = ?1 in (i), we get ? 3 = A(3) A = ?1.

Comparing coefficients of the like powers of x on both sides of (i), we get

A + B = 0, C + 2A = ?1 and C + B = 2

?1 + B = 0, C ? 2 = ?1 (Putting A = ?1)

B = 1, C = 1

2x 1

1

x 1

(x 1)(x2 2)

= ?

+ x 1

x2 2

2x 1

Hence (x 1)(x2 2) dx

= ? n |x + 1| + 1 n |x2 + 1| + 1 tan?1 x + c

2

2

2

CASE IV When some of the factors of the denominator g(x) are quadratic and repeating fractions of the

form

A0(2ax b) ax 2 bx c

ax 2

A1 bx

c

+

A1(2ax b) ax2 bx c 2

ax 2

A2 bx

c

2

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