Integral Calculus by Arihant B016

2 Integral Calculus

Do You Know It was in this aspect that the process of integration was treated by Leibnitz, the symbol of ∫ being regarded as the initial letter of the word Sum, in the same way as the symbol of differentiation d is the initial letter in the word difference.

Definition If f and g are functions of x such that g ′ ( x) = f ( x), then the function g is called a anti-derivative (or primitive function or simply integral) of f w.r.t. x. It is written symbolically, d ∫ f ( x) dx = g ( x), where dx g ( x) = f ( x)

Points to Consider 1. In other words,

∫ f ( x ) dx = g ( x ) iff

2.

g′ ( x ) = f ( x )

∫ f ( x ) dx = g ( x ) + c, where c is constant, [Q

( g ( x ) + c )′ = g′ ( x ) = f ( x )]

and c is called constant of integration.

Illustration 1 If Solution. As, ⇒ ∴

d [ x n + 1 + c ] = ( n + 1) x n , then find dx

∫x

n

dx.

d [ x n + 1 + c ] = ( n + 1) x n dx

( x n + 1 + c ) is anti-derivative or integral of ( n + 1) x n xn + 1 n ∫ x dx = n + 1 + C

Illustration 2 Evaluate ∫ e x (cos x − sin x ) dx. Solution. Here, e x (cos x − sin x ) is the derivative of e x cos x ⇒

∫e

x

(cos x − sin x ) dx = e x cos x + C

Fundamental Integration Formulae Since,

d { g ( x)} = f ( x) dx

⇔

∫ f ( x) dx = g ( x) + C

Therefore, based upon this definition and various standard differentiation formulas, we obtain the following integration formulae

Chapter 1 : Indefinite Integrals 3 (i) (ii) (iii) (iv) (v) (vi) (vii) (viii) (ix) (x) (xi) (xii) (xiii) (xiv)

d  xn + 1    = xn, n ≠ − 1 dx  n + 1 1 d (log | x |) = dx x d x x (e ) = e dx d  ax   = a x , a > 0, a ≠ 1  dx  log e a  d ( − cos x) = sin x dx d (sin x) = cos x dx d (tan x) = sec2 x dx d ( − cot x) = cosec2 x dx d (sec x) = sec x tan x dx d ( −cosec x) = cosec x cot x dx d (log |sin x |) = cot x dx d ( − log |cos x |) = tan x dx d (log |sec x + tan x |) = sec x dx d (log|cosec x − cot x|) = cosec dx d  − 1 x sin  =  dx a

(xvi)

d  −1 x  cos  =  dx a  d dx

1  a

d dx d (xix) dx

1  a

(xvii) (xviii)

(xx)

1  a

1 a − x2 −1 2

a −x 1 x tan − 1  = 2  a a + x2 −1 x cot − 1  = 2  a a + x2 x 1 sec − 1  = a x x2 − a2 2

d 1 − 1 x  cosec  = dx  a a x

2

−1 x −a 2

∫x

⇒

∫ x dx = log | x | + C, when x ≠ 0

⇒

∫e

⇒

dx =

n

1

dx = e x + C

x

x ∫ a dx =

ax +C log a

⇒

∫ sin x dx = − cos x + C

⇒

∫ cos x dx = sin x + C

⇒

∫ sec

⇒

∫ cosec

⇒

∫ sec x tan x dx = sec x + C

⇒

x dx = tan x + C

2

2

x dx = − cot x + C

∫ cosec x cot x dx = − cosec x + C

⇒

∫ cot x dx = log |sin x | + C

⇒

∫ tan x dx = − log |cos x | + C

⇒ ∫ sec x dx = log |sec x + tan x | + C x

∫ cosec x dx = log |cosec x − cot x | + C dx x = sin − 1   + C ⇒ ∫ 2 2  a a −x

⇒ (xv)

xn + 1 + C, n ≠ − 1 n+1

⇒

2

−1

x dx = cos − 1   + C  a a −x 1 dx x = tan − 1   + C 2 2  a a a +x −1 1 x dx = cot −1   + C 2 2  a a a +x

⇒

∫

⇒

∫

⇒

∫

⇒

∫x

⇒

∫x

2

2

dx x −a − dx 2

x −a 2

2

2

=

1 x sec −1   + C  a a

=

1 x cosec −1   + C  a a

4 Integral Calculus Points to Consider 1.

∫ k f ( x ) dx = k ∫ f ( x ) dx, where k is constant.

ie, The integral of the product of a constant and a function = the constant × integral of the function. 2. ∫ { f1 ( x ) ± f 2 ( x ) ± … ± f n ( x )} dx = ∫ f1 ( x ) dx ± ∫ f 2 ( x ) dx ± … ± ∫ f n ( x ) dx. ie, The integral of the sum or difference of a finite number of functions is equal to the sum or difference of the integrals of the various functions. 3. Geometrical interpretation x2 y = ∫ x dx = +C eg, 2 x2+C3 x2+C2 x2+C1

O

(x,0)

Fig. 1.1 y = ∫ f ( x ) dx = F( x ) + C

∴ ⇒

F′ ( x ) = f ( x ); F′ ( x1 ) = f ( x1 )

Hence, y = ∫ f ( x ) dx denotes a family of curves such that the slope of the tangent at x = x1 on every member is same ie, F′ ( x1 ) = f ( x ) [when x1 lies in the domain of f ( x )] Hence, anti-derivative of a function is not unique. If g1( x ) and g2( x ) are two anti-derivatives of a function f ( x ) on [ a, b], then they differ only by a constant. g1( x ) − g2( x ) = C

ie,

4. Anti-derivative of a continuous function is differentiable. ie, If f ( x ) is continuous, then ∫ f ( x ) dx = F( x ) + C ⇒ F′ ( x ) = f ( x ) ⇒ F′ ( x ) ⇒ always exists and is continuous. 5. If integrand is discontinuous at x = x1, then its anti-derivative at x = x1 need not be discontinuous. eg,

∫x

−1 / 3

dx. Here, x −1/ 3 is discontinuous at x = 0.

3 2/ 3 x + C is continuous at x = 0. 2 6. Anti-derivative of a periodic function need not be a periodic function. eg, f ( x ) = cos x + 1 is periodic but ∫ ( cos x + 1) dx = sin x + x + C is aperiodic. But

∫x

−1 / 3

dx =

Chapter 1 : Indefinite Integrals 5 Illustration 3 Evaluate (i) ∫

x 2 + 5x − 1 x

(ii) ∫ ( x 2 + 5)3 dx

dx

Solution. (i) I =

∫

=

∫

x 2 + 5x − 1 dx x  x2 5 x1 1   1/ 2 + 1/ 2 − 1/ 2  dx x x  x

− x − 1/ 2 ) dx x x − 1/ 2 + 1 5x = + − +C 3 /2 + 1 1/2 + 1 −1/2 + 1

∫ 3(/x2 + 1 + 5x

=

3/ 2

1/ 2

1/ 2 + 1

 using  I=

⇒ (ii) I =

∫ (x

2

∫x

n

dx =

 xn + 1 + C n+1 

2 5/ 2 2 x + ⋅ 5 x 3 / 2 − 2 x1 / 2 + C 5 3

+ 5)3 dx

Expanding the integrand by the binomial formula I=

∫ (x

6

+ 15x 4 + 75x 2 + 125) dx

x 7 15x 5 75x 3 + + + 125x + C 7 5 3 x7 I= + 3 x 5 + 25x 3 + 125x + C 7 I =

Illustration 4 Evaluate (i) ∫ tan 2 x dx (iii) ∫

sin 6 x + cos 6 x sin 2 x cos 2 x

Solution. (i)

dx I= I = I=

⇒ (ii)

(ii) ∫

dx sin x cos 2 x

(iv) ∫

cos x − cos 2x dx 1 − cos x

2

∫ tan x dx 2 ∫ (sec x − 1) dx 2 ∫ sec x dx − ∫ 1 dx 2

I = tan x − x + c 1 I= ∫ dx sin 2 x cos 2 x sin 2 x + cos 2 x

I =

∫

I=

∫ sin 2 x cos2 x dx + ∫ sin 2 x cos2 x dx

I=

∫ sec

sin 2 x cos 2 x

(as sin 2 x + cos 2 x = 1)

dx

sin 2 x

2

x dx +

cos 2 x

∫ cosec

I = tan x − cot x + C

2

x dx

6 Integral Calculus (iii)

I=

∫

I=

∫

sin 6 x + cos 6 x

dx

sin 2 x cos 2 x

(sin 2 x + cos 2 x )3 − 3 sin 2 x cos 2 x (sin 2 x + cos 2 x ) sin 2 x cos 2 x

dx

[ using ( a + b)3 = a 3 + b3 + 3 ab ( a + b)] I=

∫

1 − 3 sin x cos x 2

2

sin 2 x cos 2 x

dx

1

=

∫ sin 2 x cos2 x dx − ∫ 3 dx

=

∫

=

∫ sec

(sin 2 x + cos 2 x ) sin 2 x cos 2 x 2

x dx +

dx − 3 x + C

∫ cosec

2

x dx − 3 x + C

= tan x − cot x − 3 x + C cos x − cos 2x dx I= ∫ 1 − cos x

(iv)

cos x − (2 cos 2 x − 1)

I=

∫

I=

∫

I=

∫ (2 cos x + 1) dx

1 − cos x

dx

− (2 cos x + 1) (cos x − 1) − (cos x − 1)

dx

I = 2 sin x + x + C

Illustration 5 Evaluate (i) ∫

x3 dx x+2

Solution. (i)

(ii) ∫ I=

x2 dx x +5 2

x3

∫ x + 2 dx

 8  I = ∫  x 2 − 2x + 4 −  dx x + 2  x3 − x 2 + 4x − 8 log | x + 2| + C 3 x2 dx I=∫ 2 x +5 I= (ii)

 5  dx  dx = x − 5 ∫ 2 = ∫ 1 − 2 x 5 + x ( 5 )2 +   =x− ∴

5  x  tan − 1   + C  5 5

 x  I = x − 5 tan − 1   + C  5

Chapter 1 : Indefinite Integrals 7 Points to Consider In rational algebraic functions if the degree of numerator is greater than or equal to degree of denominator, then always divide the numerator by denominator and use the result of integration.

Illustration 6 Solve (i) ∫ 5log e

x

Solution. (i)

I =∫5

log e x

dx

=∫x

log e 5

dx

=

∫5

∴

(ii) ∫ 2log 4

dx

log e x

dx =

dx

(as a log e

b

= blog c a )

x log e 5 + 1 +C (log e 5 + 1) x log e 5 + 1 +C log e 5 + 1

I = ∫ 2log 4

(ii)

x

=∫2

log

x

22

dx x

= ∫ 21/ 2 log 2 = ∫ 2log 2 =∫

x

dx x

dx

1    using log bn x = log b x   n

dx

x dx

( using a log a

b

= b)

x 3/ 2 +C 3/2 2 dx = x 3/ 2 + C 3 =

∴

∫2

log 4 x

Methods of Integration If the integrand is not a derivative of a simple function, then the corresponding integrals cannot be found directly. In order to find the integral of complex problems, we follow the following rules :

Integration by Substitution (or by change of the independent variable) If g ( x) is a continuously differentiable function, then to evaluate integrals of the form, I = ∫ f ( g ( x)) . g ′ ( x) dx, we substitute

g ( x) = t and g ′ ( x) dx = dt

The substitution reduces the integral to ∫ f ( t) dt.

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

8 Integral Calculus Illustration 7 Evaluate (i) ∫ (iii) ∫

sin (log x ) x e

 3 sin x + 4 cos x  (ii) ∫   dx  4 sin x − 3 cos x 

dx

m tan −1 x

1 + x2

(iv) ∫ x sin (4x 2 + 7) dx

dx

Solution. (i) I =

sin (log x )

∫

x

dx

d 1 (log x ) = dx x

We know,

log x = t

Thus, let ∴

I=

1 dx = dt x

⇒

∫ sin ( t ) dt = − cos ( t ) + C

= − cos (log x ) + C 3 sin x + 4 cos x (ii) I = ∫ dx 4 sin x − 3 cos x d We know, (4 sin x − 3 cos x ) = (4 cos x + 3 sin x ) dx 4 sin x − 3 cos x = t

Thus, let ⇒ ∴

I=

∫

⇒

I=

∫

e

m tan − 1 x

I=

∫

I=

∫e

I=

1 m

I=

1 t 1 m tan − 1 e +C= e m m

(iv)

I=

∫ x sin (4x

Let

4x + 7 = t

⇒ ∴ ⇒

[using Eq. (i)]

…(i)

(4 cos x + 3 sin x ) dx = dt dt = log | t | + C t 3 sin x + 4 cos x dx 4 sin x − 3 cos x

= log |4 sin x − 3 cos x | + C (iii)

…(i)

dx

1 + x2 t

⋅

dt m

∫e

t

dt

2

let

[using Eq. (i)]

m tan − 1 x = t

⇒

m dx = dt 1 + x2

⇒

1 1 dx = dt m 1 + x2 x

+C

+ 7) dx

2

1 dt 8 dt I = ∫ sin ( t ) 8 1 I = − cos ( t ) + C 8 1 I = − cos (4x 2 + 7) + C 8

8x dx = dt, x dx =

Chapter 1 : Indefinite Integrals 9 Points to Consider 1

1. If

∫ f ( x ) dx = g ( x ) + C, then ∫ f ( ax + b) dx = a

2. If

∫ x dx = log | x | + C, then ∫ ax + b dx = a log | ax + b| + C

1

1

g ( ax + b) + C

1

Thus, in any fundamental integral formulae given in article fundamental integration formulae if in place of x we have ( ax + b), then same formula is applicable but we must divide by coefficient of x or derivative of (ax + b) ie, a.

Here is the list of some of frequently used formulae (i) (ii)

n ∫ ( ax + b) dx =

1

ax + b

(iv)

bx + c dx = ∫a

(vii) (viii) (ix) (x) (xi) (xii) (xiii) (xiv)

+ C, n ≠ − 1

1

∫e

(vi)

a ( n + 1)

∫ ax + b dx = a log | ax + b| + C

(iii)

(v)

( ax + b) n + 1

dx =

1 ax + b e +C a 1 a bx + c ⋅ +C b log a 1

∫ sin ( ax + b) dx = − a cos ( ax + b) + C 1

∫ cos ( ax + b) dx = a sin ( ax + b) + C 1 tan ( ax + b) + C a 1 2 ∫ cosec ( ax + b) dx = − a cot ( ax + b) + C 1 ∫ sec ( ax + b) tan ( ax + b) dx = a sec ( ax + b) + C 1 ∫ cosec ( ax + b) cot ( ax + b) dx = − a cosec ( ax + b) + C 1 ∫ tan ( ax + b) dx = − a log |cos ( ax + b) | + C 1 ∫ cot ( ax + b) dx = a log |sin ( ax + b) | + C 1 ∫ sec ( ax + b) dx = a log |sec ( ax + b) + tan ( ax + b) | + C 1 ∫ cosec ( ax + b) dx = a log |cosec ( ax + b) − cot ( ax + b) | + C

∫ sec

2

( ax + b) dx =

10 Integral Calculus Illustration 8 Evaluate (i) ∫ cos 4x cos 7x dx

(ii) ∫ cos x cos 2x cos 5x dx

Solution. When calculating such integrals it is advisable to use the trigonometric product formulae. (i)

∫ cos 4x cos 7x dx

Here, cos 4x cos 7x = ∴

(ii)

I=

1 (cos 3 x + cos 11x ) 2

∫ cos 4x cos 7x dx

=

1 2

∫ (cos 3 x + cos 11x ) dx

=

1 2

∫ cos 3 x dx + 2 ∫ cos 11x dx

=

sin 3 x

1

6

+

sin 11x 22

∫ cos x cos 2x cos 5x dx

We have, (cos x cos 2x ) cos 5x =

+C

1 (cos x + cos 3 x ) cos 5x 2

=

1 {2 cos x cos 5x + 2 cos 3 x cos 5x} 4

=

1 { (cos 4x + cos 6x ) + (cos 2x + cos 8x )} 4

∴

cos x cos 2x cos 5x =

∴

I=

1 {cos 2x + cos 4x + cos 6x + cos 8x} 4

∫ (cos x cos 2x cos 5x ) dx

1 (cos 2x + cos 4x + cos 6x + cos 8x ) dx 4∫ sin 2x sin 4x sin 6x sin 8x = + + + +C 8 16 24 32

=

Points to Consider While solving such problems it is expedient to use the following trigonometric identities : 1 1. sin mx cos nx = {sin ( m − n ) x + sin ( m + n ) x } 2 1 2. cos mx sin nx = {sin ( m + n ) x − sin ( m − n ) x } 2 1 3. sin mx sin nx = {cos ( m − n ) x − cos ( m + n ) x } 2 1 4. cos mx cos nx = {cos ( m − n ) x + cos ( m + n ) x } 2

Chapter 1 : Indefinite Integrals 11 Illustration 9 Evaluate (i) ∫ x (iii) ∫ (v)

x − 5 dx

(ii) ∫

x3 dx 1 + 2x

(iv) ∫

x2

∫ ( a + bx )2 dx

Let

(ii) I = ∫ Let

∫x

x −5 = t

∴

I=

8x + 13 4x + 7

4x + 7 x2 + 1 ( x 2 − 1 )2

(vi) ∫ I=

Solution. (i)

8x + 13

dx

dx x+1− x

x − 5 dx ⇒

2

∫ (t

dx

2

dx = 2t dt

+ 5) ⋅ t ⋅ 2t dt = 2 ∫ ( t 4 + 5t 2 ) dt

 t 5 5t 3  =2  + +C 3   5  ( x − 5 )5 / 2 5 ( x − 5 )3 / 2  =2  + +C 5 3   dx

4x + 7 = t 2 1 t dt 2 2 2 ( t − 7) + 13 1 ⋅ t dt t 2

4dx = 2t dt, dx =

∴

∫

∴

I=

⇒

1 (2t 2 − 14 + 13 ) dt I= 2 ∫ 1 (2t 2 − 1) dt = 2 ∫ 1 t I = t3 − + C 3 2 1 1 I = (4x + 7)3/ 2 − (4x + 7)1/ 2 + C 3 2

∴ (iii) I = ∫

x3 dx, let 1 + 2x = t 2 1 + 2x 2dx = 2t dt, dx = t dt 3

∴

∴

 t2 − 1    ⋅ t dt  2 

∫ t 1 6 ( t − 3 t 4 + 3 t 2 − 1) dt I= 8 ∫  1  t7 3 =  − t5 + t3 − t  + C 87 5  I=

I=

7/ 2  1  (2x + 1) 3 − (1 + 2x )5/ 2 + (2x + 1)3/ 2 − (2x + 1)1/ 2  + C  8 7 5 

12 Integral Calculus (iv)

∴

(v)

x2 + 1

I=

∫ ( x 2 − 1)2 dx

I=

∫

1 + 1 / x2

(dividing Nr and Dr by x 2) dx 2 1  x−   x  dt 1 1   I = ∫ 2 let x − = t ⇒  1 + 2  dx = dt  x x  t 1 =− +C t 1 x +C= +C I=− 1  1 − x2 − x    x 

I= = I=

x2

 2a a2  x− 2   − b b  

1

∫ ( a + bx )2 dx = ∫ b2 + 1 b2

dx

( bx + a )2

2bx + a

a

∫ 1 dx − b2 ∫ ( bx + a )2 dx

1 a  x− 2  2 b  b

∫

2bx + 2a − a ( bx + a )

2

 dx  

1 a  1 dx − a x − 2 2∫ 2 ( bx + a ) b  b

=

 a 1 a 2 x − 2  log | bx + a | + +C b ( bx + a )  b  b b2

∫ ( bx + a )

−2

 1  a2 bx − 2a log | bx + a | − +C 3  ( a + bx )  b  dx (rationalising Dr ) I= ∫ x+1− x =

(vi)

 dx  

=

x+1+

I=

∫

I=

∫(

I=

x

( x + 1) − ( x ) x+1+

dx

x ) dx

2 2 ( x + 1 )3 / 2 + ( x )3 / 2 + C 3 3

Integral of the Form 1  1 1. ∫ f  x +  1 − 2  dx    x x  1 1  Put x + = t ⇒ 1 − 2  dx =  x x  1 1  1 2. ∫ f  x −  1 + 2  dx Put x − = t ⇒  x  x x 

dt 1  1 + 2  dx = dt  x 

Chapter 1 : Indefinite Integrals 13 3. ∫

x2 + 1 x4 + kx2 + 1

dx

Divide numerator and denominator by x2 x2 − 1 4. ∫ 4 dx x + kx2 + 1 Divide numerator and denominator by x2 Illustration 10 Solution. I = ∫ Put x x = y

y−

x

x x ( x 2x + 1) (ln x + 1)

x (x

2x

x 4x + 1 + 1) (ln x + 1) x 4x + 1

dx

dx

⇒ x x (ln x + 1) dx = dy I=∫

Let

∫

y2 + 1 y4 + 1

1 1 1+ 2 y y2 dy = ∫ dy 2 1  1 y2 + 2 y−  + 2 y  y 1+

dy = ∫

 1 1 + 2  dy = dt  y  1  dt −1 t  I=∫ 2 =  +C  tan  2 2 t +2 1  y−  1 1 y tan −1  tan −1 +C= = 2 2  2     

1 =t ⇒ y

Illustration 11 Evaluate

∫

1  x x − x x  +C  2    

( x 2 − 1) dx 1  ( x 4 + 3 x 2 + 1) tan − 1  x +   x 

.

Solution. The given integral can be written as I=

I=

Let ∴

∫ ∫

(1 − 1 / x 2 ) dx ( x + 3 + 1 / x ) tan 2

2

−1

1   x+   x 

(dividing Nr and Dr by x 2)

(1 − 1 / x 2 ) dx 1  {( x + 1 / x )2 + 1} tan − 1  x +   x  1 1   x + = t ⇒  1 − 2  dx = dt  x x  dt I= ∫ 2 ( t + 1) ⋅ tan − 1 ( t )

Now, make one more substitution tan − 1 t = u. Then,

dt = du t +1 2

…(i)

14 Integral Calculus du = log |u | + C u

∫

∴ Eq. (i) becomes,

I=

⇒

I = log |tan − 1 t | + C = log |tan − 1 ( x + 1 / x )| + C ( x − 7/ 6 − x 5/ 6 ) dx

∫ x1/ 3 ( x 2 + x + 1)1/ 2 − x1/ 2 ( x 2 + x + 1)1/ 3 ⋅

Illustration 12 Solution. I = ∫ =∫

=∫

x 7/ 6 ( x − 7/ 6 − x 5/ 6 ) dx x 7/ 6 ⋅ x1/ 3 ( x 2 + x + 1)1/ 2 − x1/ 2 ⋅ x 7/ 6 ( x 2 + x + 1)1/ 3 (1 − x 2 ) dx x

3/ 2

1   −  1 − 2  dx  x  1    x+ +1   x

=−∫ Substitute,

( x + x + 1)1/ 2 − x 5/ 3 ( x 2 + x + 1)1/ 3 2

1/ 2

1   − x+ +1   x

dt ( t + 1)1/ 2 − ( t + 1)1/ 3

1/ 3

1  Putting x+ =t  x  1    ⇒  1 − 2  dx = dt   x 

( t + 1) = u 6 =−∫

6u 5 du u3 − u2

= −6 ∫ = −6 ∫ = −6 ∫

u3 du, put u − 1 = z u −1 ( z + 1 )3 dz z z3 + 3 z2 + 3 z + 1 z

dz

1  = − 6 ∫  z 2 + 3 z + 3 +  dz  z

where,

 z3 3 z2  = −6  + + 3 z + log | z |  + C 2  3  1/ 6 1   z= x+ +1 −1   x

Illustration 13 The value of ∫ {{[ x ]}} dx, where {.} and [.] denotes fractional part of x and greatest integer function) is equal to (a) 0

(b) 1

(c) 2

(d) –1

Solution. Let I = ∫ {{[ x ]}} dx where, ∴

[ x ] = Integer and we know {n} = 0; n ∈ Integer. I = ∫ 0 dx = 0

Hence, (a) is the correct answer.

Chapter 1 : Indefinite Integrals 15 Illustration 14 The value of ∫ [{ x }] dx; (where [.] and {.} denotes greatest integer and fractional part of x is equal to (a) 0

(b) 1

(c) 2

(d) –1

Solution. As, we know y = { x } could be shown as y

1 x

O

∴

[{ x }] = 0 I = ∫ [{ x }] dx = ∫ 0 dx = 0

Thus,

Hence, (a) is the correct answer.

Illustration 15 The value of (a) 2 x 2 + 2 + C

Solution. Here, We know, ∴

∫

d ( x 2 + 1)

(b) x 2 + 2 + C d ( x 2 + 1) I=∫ x2 + 2

x2 + 2 (c) x

x2 + 2 + C

(d) None of these

d ( x 2 + 1) = 2x dx 2x dx I=∫ x2 + 2

Put,

x2 + 2 = t2

∴

2x dx = 2t dt 2t dt = 2t + C I=∫ t

⇒

, is

I = 2 x2 + 2 + C

Hence, (a) is the correct answer.

Illustration 16 If ∫ (a) 2 / 5, 5 / 2

 xk  ( x )5   + c, then a and k are = dx a log k ( x )7 + x 6  1+ x 

(b) 1 / 5, 2 / 5 (c) 5 / 2, 1 / 2 ( x )5 Solution. Here, I = ∫ dx ( x )7 + x 6 =∫ =∫

(d) 2 / 5, 1 / 2

dx ( x )2 + ( x )7 dx 1 5 = y ⇒ − 7/ 2 dx = dy , put 1  x 5/ 2 2x 7/ 2  x  1 + 5/ 2  x  

16 Integral Calculus =− =

2 5

dy

∫1+

y

=−

2 log |1 + y| + C 5

 1  2 + C log  5 1 + y

 x 5/ 2  2 + C log  5/ 2 5 + 1 x  xk   +C I = a log  k  1+ x  =

where,

…(i) (given)

…(ii)

∴ From Eqs. (i) and (ii), we get  xk   x 5/ 2  2    +C log + = C a log  5/ 2 k 5  1+ x   1+ x  ⇒

a =2/5

and

k =5/2

Hence, (a) is the correct answer.

Illustration 17 The value of

∫

1 − 2 cos 3 x (b) sin x −

(a) sin x + sin 2x + C (c) − sin x −

cos 5x + cos 4x

sin 2x +C 2

Solution. Here, I = ∫

dx, is equal to

sin 2x 2

+C

(d) None of these cos 5x + cos 4x dx 1 − 2 cos 3 x

x 9x ⋅ cos 2 2 dx =∫ 3x   −1  1 − 2  2 cos 2   2 9x x 2 cos cos 2 2 dx =∫ 3x 3 − 4 cos 2 2 3x Multiplying and dividing it by cos , we get 2 3x 9x x 2 cos ⋅ cos ⋅ cos 2 2 2 dx =∫ 3x 3 3x 3 cos − 4 cos 2 2 2 cos x / 2 ⋅ cos 3 x / 2 ⋅ cos 9x / 2 =∫ dx − cos 9x / 2 2 cos

(using cos 3θ = 4 cos 3 θ − 3 cos θ ) 3x x = − ∫ 2 cos cos 2 2  sin 2x =− + sin x  2 Hence, (c) is the correct answer.

dx = − ∫ (cos 2x + cos x ) dx   +C 

Chapter 1 : Indefinite Integrals 17 Illustration 18 The value of cos 3 x + +C 2 3 sin 2x cos 3 x (c) − +C 2 3

(a)

sin 2x

∫

cos 7x − cos 8x 1 + 2 cos 5x

dx, is equal to

(b) sin x − cos x + C (d) None of these

Solution. Here, I = ∫ =∫

cos 7x − cos 8x 1 + 2 cos 5x

dx

15x x ⋅ sin 2 2 dx 1 + 2 cos 5x

2 sin

5x , 2 15x 5x x 2 sin ⋅ sin ⋅ sin 2 2 2 dx =∫ 5x 5x + 2 sin ⋅ cos 5x sin 2 2 x 15x 5x 2 sin ⋅ sin ⋅ sin 2 2 2 dx =∫ 5x 15 5x sin x − sin + sin 2 2 2 5x x = ∫ 2 sin ⋅ sin dx 2 2

Multiplying and dividing by sin

= ∫ (cos 2x − cos 3 x ) dx I=

sin 2x sin 3 x − +C 2 3

Hence, (c) is the correct answer.

Illustration 19 If

1

∫ f ( x ) cos x dx = 2 f

(a) x

(b) 1

Solution. Here,

∫

2

(c) cos x 1 2 f ( x ) cos x dx = f ( x ) + C 2

Differentiating both the sides, we get

ie, ⇒ ⇒

( x ) + c, then f ( x ) can be

f ( x ) cos x = f ( x ) ⋅ f ′ ( x ) d cos x = ( f ( x )) dx f ( x ) = ∫ cos x dx f ( x ) = sin x + C

Hence, (d) is the correct answer.

(d) sin x

18 Integral Calculus

Target Exercise 1.1 Solved the following integration : 2 + 3x2 1. ∫ 2 dx x (1 + x 2 ) 3.

x2 + 3

∫ x 6 ( x 2 + 1)

2.

dx

(1 + x )2

5.

∫ x (1 + x 2 ) dx

7.

∫ x 2 + 1 dx

9.

∫

x6 − 1

∫

dx x+1− x 1 + 2x 2

4.

∫ x 2 (1 + x 2 ) dx

6.

∫ 1 + x 2 dx

8.

∫

x4

x4 + x2 + 1 2 (1 + x 2 )

dx

( x + 1) ( x 2 − x ) dx x x +x+ x 1 − x −2



x −2 − x

2



10.

∫  x1/ 2 − x −1/ 2 − x 3/ 2 + x1/ 2 − x −1/ 2  dx

11.

∫  4 + 2x −1 + x −2 ⋅ 4 − 4x −1 + x −2 −

12.

∫

14.

∫

16.

∫

18.

∫ 1 + cos 2x dx

20.

∫ cos2 x sin 2 x dx

22.

∫ cos x + sin x (2 + 2 sin 2x ) dx

23. ∫ (3 sin x cos 2 x − sin 3 x ) dx

24.

∫ cos x ° dx

25.

∫ sec 2x + 1 dx

26.

∫

27.

∫

x −6 − 64



x2

( x 2 + sin 2 x ) sec 2x 1 + x2 e 3x + e 5x e x + e −x dx 1 + sin x

4x 2 (2x + 1)  dx 1 − 2x 

dx

15. ∫ ( e a ln x + e x ln a ) dx

dx

17. ∫ sin x cos x cos 2x cos 4x dx

1 + cos 2 x

19.

cos 2x

cos x − sin x

cos x − cos 2x 1 − cos x

∫

30.

∫ sec

32.

∫

33.

∫

dx

cos 2x − cos 2α

28.

cos x − cos α 2

13. ∫ 2x ⋅ e x dx

dx

x cosec 2x dx

1 − tan 2 x

∫ 1 + tan 2 x dx

21. ∫ 4 cos

21 x ⋅ cos x ⋅ sin x dx 2 2

sec 2x − 1

sin x + cos x dx (cos x + sin x > 0) 1 + sin 2x sin 3 x + cos 3 x

29.

∫

31.

∫

34.

∫ cot x − tan x dx

sin 2 x cos 2 x

dx

1 − sin 2x dx

sin x + cos x 6

6

dx sin 2 x ⋅ cos 2 x  2  9 π x x  2  7π sin  8 + 4  − sin  8 + 4   dx  

cos 4x + 1

Chapter 1 : Indefinite Integrals 19 35. 37.



∫ sin α sin ( x − α ) + sin ∫

2

 x − α   dx 2  

cot2 2x − 1  − cos 8x cot 4x  dx   2 cot 2x 

36. 38.

∫ ∫

sin 2x + sin 5x − sin 3 x cos x + 1 − 2 sin 2 2x cos 4 x − sin 4 x 1 + cos 4x

dx

dx (cos 2x > 0)

Integration by Parts Theorem

If u and v are two functions of x, then   du ∫ uv dx = u ∫ v dx − ∫  dx ∫ v dx dx

ie, The integral of product of two functions = (first function) × (integral of second function) – integral of (differential of first function × integral of second function). Proof

∴

For any two functions f ( x) and g( x), we have d d d { f ( x) ⋅ g ( x)} = f ( x) ⋅ { g ( x)} + g ( x) ⋅ { f ( x)} dx dx dx d d   ∫  f ( x) ⋅ dx { g ( x)} + g ( x) ⋅ dx { f ( x)} dx = ∫ f ( x) ⋅ g( x) dx 

d



d





d



or

∫  f ( x) ⋅ dx { g( x)} dx + ∫  g ( x) ⋅ dx { f ( x)} dx = ∫ f ( x) ⋅ g ( x) dx

or

∫  f ( x) ⋅ dx { g ( x)}  dx = ∫ f ( x) ⋅ g ( x) dx − ∫  g ( x) ⋅ dx { f ( x)} dx

Let

f ( x) = u and



d



d { g ( x)} = v dx

So that g ( x) = ∫ v dx ∴



  du ⋅ v dx ⋅ dx dx ∫ 

∫ uv dx = u ⋅ ∫ v dx − ∫ 

Points to Consider While applying the above rule, care has to be taken in the selection of first function (u) and selection of second function (v). Normally we use the following methods : 1. If in the product of the two functions, one of the functions is not directly integrable (eg, log | x |, sin − 1 x, cos − 1 x, tan − 1 x, … , etc.) Then, we take it as the first function and the remaining function is taken as the second function. eg, In the integration of ∫ x tan − 1 x dx, tan − 1 x is taken as the first function and x as the second function. 2. If there is no other function, then unity is taken as the second function. eg, In the integration of ∫ tan − 1 x dx, tan − 1 x is taken as first function and 1 as the second function. 3. If both of the function are directly integrable, then the first function is chosen in such a way that the derivative of the function thus obtained under integral sign is easily integrable.

20 Integral Calculus Usually we use the following preference order for selecting the first function. (Inverse, Logarithmic, Algebraic, Trigonometric, Exponent). In above stated order, the function on the left is always chosen as the first function. This rule is called as ILATE. Illustration 20 Evaluate (i) ∫ sin − 1 x dx

(ii) ∫ log e | x | dx

Solution. (i) I =

∫ sin

−1

∫ sin

x dx =

−1

I

x ⋅ 1 dx II

Here, we know by definition of integration by parts that order of preference is taken according to ILATE. So, ‘ sin − 1 x ’ should be taken as first and ‘1’ as the second function to apply by parts. Applying integration by parts, we get 1 I = sin − 1 x ⋅ ( x ) − ∫ ⋅ x dx 1 − x2 = x ⋅ sin − 1 x +

= x sin

−1

1 2

dt

∫ t1/ 2

let

1 − x2 = t − 2x dx = dt 1 x dx = − dt 2

1 t1/ 2 x+ ⋅ +C 2 1/2

I = x sin − 1 x + 1 − x 2 + C ∴

∫ sin

(ii)

I=

−1

x dx = x sin − 1 x + 1 − x 2 + C

∫ log e | x | dx = ∫ log e | x |⋅ 1 dx

I Applying integration by parts, we get 1 = log | x |⋅ x − ∫ ⋅ x dx x

II

= x log | x | − ∫ 1 dx

I = x log | x | − x + C

Illustration 21 Evaluate (i) ∫ x cos x dx

Solution. (i) I=

∫ x cos x dx

(ii) ∫ x 2 cos x dx

∫ x cos x dx

I II Applying integration by parts,  d I = x ( ∫ cos x dx ) − ∫  ( x ) { ∫ (cos x ) dx } dx   dx I = x sin x −

∫ 1 ⋅ sin x dx

I = x sin x + cos x + C

Chapter 1 : Indefinite Integrals 21 (ii)

I=

∫x

2

cos x dx I II Applying integration by parts, d

I = x 2 ( ∫ cos x dx ) −

∫  dx ( x

2

 ) ⋅ { ∫ cos x } dx 

= x 2 sin x − ∫ 2x ⋅ (sin x ) dx = x 2 sin x − 2 ∫ x (sin x ) dx We again have to integrate ∫ x sin x dx using integration by parts,

∫ x ⋅ sin x dx

= x 2 ⋅ sin x − 2

I

II

   dx  = x 2 sin x − 2 x ( ∫ sin x dx ) − ∫   ( ∫ sin x dx ) dx   dx   

∫ 1 ⋅ ( − cos x ) dx }

= x 2 sin x − 2 { − x cos x −

I = x sin x + 2x cos x − 2 sin x + C 2

Illustration 22 Evaluate Solution.

sin − 1

∫ sin − 1

sin − 1

∫ sin − 1

x − cos − 1 x + cos

x

−1

x

x − cos − 1 x + cos

dx =

∫

−1

sin − 1

x x

dx.

x − ( π / 2 − sin − 1 π/2

x)

dx

(Q sin − 1 θ + cos − 1 θ = π / 2) ⇒

Let ∴

2 π 4 I= π 4 I= π I=

∫ (2 sin

−1

x − π / 2) dx

∫ sin

−1

x dx −

∫ sin

−1

x dx − x + C

∫ 1 dx …(i)

x = sin 2 θ, then dx = 2 sin θ cos θ dθ = sin 2θ dθ

∫ sin

−1

x dx =

∫ θ ⋅ sin 2θ dθ I

II

Applying integration by parts cos 2θ −θ 1 1 −1 ∫ sin x dx = − θ ⋅ 2 + ∫ 2 cos 2θ dθ = 2 ⋅ cos 2θ + 4 sin 2θ −1⋅θ 1 = ⋅ (1 − 2 sin 2 θ ) + ⋅ sin θ ⋅ 1 − sin 2 θ 2 2 −1 1 …(ii) = sin − 1 x (1 − 2x ) + ⋅ x 1 − x 2 2 From Eqs. (i) and (ii), we get 1 4 − 1 (sin − 1 x ) (1 − 2x ) + I=  π  2 2 =

2 { x − x 2 − (1 − 2x ) sin − 1 π

x

 1 − x − x + C 

x} − x + C

22 Integral Calculus Integral of the Form ∫ e x { f ( x) + f ′ ( x)} dx Theorem Prove that ∫ e x { f ( x ) + f ′ ( x )} dx = e x f ( x ) + C Proof

We have,

∫e

x

=

{ f ( x ) + f ′ ( x )} dx

∫e

∫e

⋅ f ( x ) dx + I II x

x

⋅ f ′ ( x ) dx

= f ( x ) ⋅ e x − ∫ f ′ ( x ) ⋅ e x dx + = f (x) ⋅ e + C x

∫e

x

⋅ f ′ ( x ) dx + C

Thus, to evaluate the integrals of the type ∫ e x { f ( x) + f ′ ( x) } dx, we first express the integral as the sum of two integrals ∫ e x f ( x) dx and ∫ e x f ′ ( x) dx and then integrate the integral involving e x f ( x) as integrand by parts taking e x as second function.

Points to Consider The above theorem is also true, if we have e kx in place of e x . ie, kx kx ∫ e { f ( kx ) + f ′ ( kx )} dx = e f ( kx ) + C

General Concept

∫e

g(x)

{ f ( x) g ′ ( x) + f ′ ( x)} dx

Proof

I =

∫ e{

g(x)

f ( x) g ′ ( x) dx + { 123

II

I

= f ( x) ⋅ e g(x) − eg, ⇒ ⇒ ⇒ eg,

II

∫ f ′ ( x) ⋅ e

∫e

g (x)

g(x)

f ′ ( x) dx

dx +

∫e

g(x)

⋅ f ′ ( x) dx = f ( x) ⋅ e g(x)

2   2 (x sin x + cos x) x cos x − ( x sin x + cos x)  dx  e ∫  2 x   sin + cos x x x  (x sin x + cos x)  2  dx cos x − ∫e   x2 cos cos x x   +  ′ dx (x sin x + cos x)      x cos x  ∫e  x   x  cos x e (x sin x + cos x) ⋅ +C x tan x tan x tan x ∫ e (sin x − sec x) dx = ∫ e sin x dx − ∫ e sec x dx

⇒

− e tan x ⋅ cos x +

⇒

−e

tan x

∫e

tan x

sec2 x cos x dx −

∫e

tan x

⋅ cos x

Illustration 23 Evaluate  1 + sin x cos x   dx (i) ∫ e x  cos 2 x  

 1 + sin 2x  (ii) ∫ e 2x   dx  1 + cos 2x 

sec x dx

Chapter 1 : Indefinite Integrals 23  1 + sin x cos x   dx cos 2 x  

Solution. (i) I = ∫ e x 

 1 sin x cos x  I = ∫ ex  2 +  dx cos 2 x  cos x

∫ e {tan x + sec x} dx I = ∫ e x ⋅ tan x dx + ∫ e x (sec 2 x ) dx II I=

2

x

I

I = tan x ⋅ e − ∫ sec 2 x ⋅ e x dx + ∫ e x ⋅ sec 2x dx + C x

(ii)

I = e x tan x + C   1 + sin 2x  2 x 1 + 2 sin x cos x I = ∫ e 2x   dx  dx = ∫ e  2 2 cos x 1 + cos 2x     1 2 sin x cos x   2 x 1 2 = ∫ e 2x  +  dx = ∫ e  sec x + tan x  dx 2 2 2 2 cos 2 cos x x     1 = ∫ e 2x ⋅ tan x dx + ∫ e 2x ⋅ sec 2 x dx 2 II I = tan x ⋅ I=

e 2x e 2x 1 − ∫ sec 2 x ⋅ dx + 2 2 2

∫e

2x

⋅ sec 2 x dx

1 2x e ⋅ tan x + C 2 2

 1−x   dx. 2 1 + x 

Illustration 24 Evaluate ∫ e x  2

Solution. I =

∫

ex

(1 − 2x + x 2 ) (1 + x 2 )2

dx

  1 + x2 2x − dx  2 2 2 2 (1 + x )  (1 + x )     1 d  1  2x 2x   =− = ∫ ex  − dx as 2 2 2 2 2 2 (1 + x )  (1 + x )  1 + x  dx  1 + x  x e = +C 1 + x2 ex +C I= 1 + x2 I=

∴

∫

 1−x   dx = e  2 1 + x  x

∫e

x

Integrals of the Form ∫ e ax sin bx dx , ∫ e ax cos bx dx Let Then,

∫ e (sin bx) dx I = ∫ sin bx ⋅ e ax dx

I =

ax

I

II

e  I = sin bx ⋅   −  a  ax

∫ b cos bx ⋅

e ax dx a

24 Integral Calculus  1 e ax e ax b sin bx ⋅ e ax − cos bx ⋅ dx − ∫ ( − b sin bx) ⋅ a a a  a  2 b b 1 = sin bx ⋅ e ax − 2 cos bx ⋅ e ax − 2 ∫ sin bx ⋅ e ax dx a a a 1 b b2 I = sin bx ⋅ e ax − 2 cos bx ⋅ e ax − 2 I a a a 2 ax 1⋅ e b ⋅ ( a sin bx − b cos bx) I + 2 I = a2 a  a2 + b2  e ax  = 2 ( a sin bx − b cos bx) I  2 a   a I =

∴ ⇒

e ax ( a sin bx − b cos bx) + C a2 + b2

or

I =

Thus,

∫e

ax

sin bx dx =

e ax ( a sin bx − b cos bx) + C a2 + b2

Similarly,

∫e

ax

cos bx dx =

e ax ( a cos bx + b sin bx) + C a + b2

Aliter

Use Euler’s equation

∫ e cos bx dx and Q = ∫ e P + iQ = ∫ e ax ⋅ e ibx dx = ∫ e (a + ib) x dx P=

Let Hence,

2

ax

P + iQ = = P=

∴

Q=

ax

sin bx dx

a − ib ax 1 e (cos bx + i sin bx) e (a + ib) x = 2 a + ib a + b2

( ae ax cos bx + be ax sin bx) − i ( ae ax sin bx − be ax cos bx) a2 + b2 e ax ( a cos bx + b sin bx) a2 + b2 e ax ( a sin bx − b cos bx) a2 + b2

Illustration 25 Evaluate (i) ∫ e x cos 2 x dx

(ii) ∫ sin (log x ) dx 1 + cos 2x   dx 2  

Solution. (i) I = ∫ e x ⋅ cos 2 x dx = ∫ e x ⋅  1 1 e x dx + 2∫ 2 1 x 1 I = e + I1 2 2 I=

where

∫ cos 2x ⋅ e

I1 = ∫ cos 2x ⋅ e x dx

x

dx …(i)

Chapter 1 : Indefinite Integrals 25 I1 = ∫ cos 2x ⋅ e x dx = cos 2x ⋅ e x − ∫ − 2 sin 2x ⋅ e x dx II I x = e ⋅ cos 2x + 2 ∫ sin 2x ⋅ e x dx II I = e x ⋅ cos 2x + 2 {sin 2x ⋅ e x − ∫ 2 cos 2x ⋅ e x dx }

∴

= e x ⋅ cos 2x + 2 sin 2x ⋅ e x − 4 I1 1 I1 = {e x cos 2x + 2 sin 2x e x} 5

From Eqs. (i) and (ii), we get 1 I = ex + 2 1 I = ex + 2

1 1 x ⋅ {e cos 2x + 2 sin 2x ⋅ e x} 2 5 1 x e { cos 2x + 2 sin 2x } + C 10

I = ∫ sin (log x ) dx

(ii) Let

…(ii)

log x = t

⇒

x = e t or dx = e t dt

I = ∫ (sin t ) ⋅ e t dt = sin t ⋅ e t − ∫ cos t ⋅ e t dt I II I II

∴

I = sin t ⋅ e t − {cos t ⋅ e t − ∫ − sin t ⋅ e t dt } I = e t ⋅ sin t − e t ⋅ cos t − I 1 I = e t (sin t − cos t ) + C 2 x I = { sin (log x ) − cos (log x )} + C 2

∴

∫

Illustration 26 Evaluate Solution. Let I = ∫

x 2dx ⋅ ( x sin x + cos x )2

x2 dx ( x sin x + cos x )2

Multiplying and dividing it by ( x cos x ), we get ( x cos x ) dx I = ∫ ( x sec x ) ⋅ ( x sin x + cos x )2 I II I = x sec x ⋅ ∫

= x sec x ⋅

x cos x ( x sin x + cos x )2

dx

 x cos x d  ( x sec x ) ∫ dx  dx −∫ 2  dx   ( x sin x + cos x )  −1 ( x sin x + cos x ) −1 dx ( x sin x + cos x ) ( x sin x + cos x ) dx cos 2 x ⋅ ( x sin x + cos x )

− ∫ ( x sec x ⋅ tan x + sec x ) ⋅ =

− x sec x + ( x sin x + cos x )

∫

26 Integral Calculus − x sec x + sec 2 x dx ( x sin x + cos x ) ∫ − x sec x I= + tan x + C ( x sin x + cos x ) =

Integration of the Type ∫ (sin m x ⋅ cos n x) dx (i) Where m, n belongs to natural number. (ii) If one of them is odd, then substitute for term of even power. (iii) If both are odd, substitute either of them. (iv) If both are even, use trigonometric identities only.  m + n − 2 (v) If m and n are rational numbers and   is a negative   2 integer, then substitute cot x = p or tan x = p which so ever is found suitable. Illustration 27 Evaluate ∫ sin 3 x ⋅ cos 5 x dx.

Solution. I = ∫ sin 3 x ⋅ cos 5 x dx

cos x = t ⇒ − sin x dx = dt 2 5 I = − ∫ (1 − t ) ⋅ t dt t8 t6 I = ∫ t 7 dt − ∫ t 5 dt = − +C 8 6 8 6 cos x cos x I= − +C 8 6 I = ∫ R3 (1 − R2 )2 dR, if sin x = R, cos x dx = dR

Let

Aliter

I = ∫ R3 dR − ∫ 2 R5 dR + I=

sin 4 x 4

−

2 sin 6 x 6

+

∫ R dR 7

sin 8 x 8

+C

Points to Consider This problem can also be handled by successive reduction or by trigonometrical identities. Answers will be in different form but identical with modified constant of integration.

Illustration 28 Evaluate ∫ sin − 11/ 3 x ⋅ cos − 1/ 3 x dx. Solution. Here, ∫ sin ∴

I=∫

− 11 / 3

cos − 1/ 3 x sin

− 1/ 3

x ⋅ sin 4 x

x ⋅ cos

− 1/ 3

x dx

ie,

  11 1 − − 2 − 3 3  = −3  2    

dx = ∫ (cot− 1/ 3 x ) (cosec 2 x )2 dx

Chapter 1 : Indefinite Integrals 27 I = ∫ (cot− 1/ 3 x ) (1 + cot2 x ) cosec 2 x dx (let cot x = t, − cosec 2 x dx = dt) = − ∫ t − 1/ 3 (1 + t 2 ) dt = − ∫ ( t − 1/ 3 + t 5/ 3 ) dt

3 3 3  3  = −  t 2/ 3 + t 8 / 3  + C = −  (cot2/ 3 x ) + (cot8 / 3 x ) + C 8 8 2  2 

Illustration 29 Evaluate (i) ∫

1 dx sin ( x − a ) cos ( x − b)

Solution. (i) I = ∫

(ii) ∫

1 dx cos ( x − a ) cos ( x − b)

1 dx sin ( x − a ) cos ( x − b)

cos ( a − b) dx ⋅ cos ( a − b) ∫ sin ( x − a ) cos ( x − b) cos {( x − b) − ( x − a )} 1 = ⋅ dx cos ( a − b) ∫ sin ( x − a ) cos ( x − b)

I=

cos ( x − b) ⋅ cos ( x − a ) sin ( x − b) ⋅ sin ( x − a )  1 + ⋅   dx sin ( x − a ) cos ( x − b)  cos ( a − b) ∫  sin ( x − a ) cos ( x − b) 1 {cot ( x − a ) + tan ( x − b)} dx = cos ( a − b) ∫ 1 = {log |sin ( x − a )| − log |cos ( x − b)|} + C cos ( a − b) =

 sin ( x − a )  1 + C log e  cos ( a − b)  cos ( x − b)  1 I=∫ dx cos ( x − a ) cos ( x − b) sin ( a − b) 1 dx = ∫ sin ( a − b) cos ( x − a ) cos ( x − b) sin { ( x − b) − ( x − a )} 1 = dx sin ( a − b) ∫ cos ( x − a ) cos ( x − b) =

(ii)

sin ( x − b) cos ( x − a ) cos ( x − b) sin ( x − a )  1 −  dx  sin ( a − b) ∫ cos ( x − a ) cos ( x − b) cos ( x − a ) cos ( x − b)  1 {tan ( x − b) − tan ( x − a )} dx = sin ( a − b) ∫ 1 = [ − log |cos ( x − b)| + log |cos ( x − a )|] + C sin ( a − b)

=

=

  cos ( x − a )  1  + C log  sin ( a − b)   cos ( x − b) 

Illustration 30 Evaluate Solution. Let

I=∫

sin ( x + a )

∫ sin ( x + b) dx.

sin ( x + a ) dx sin ( x + b)

Put x + b = t ⇒ dx = dt

28 Integral Calculus sin ( t − b + a )

I=∫

∴

sin t

dt =

sin t cos ( a − b) cos t sin ( a − b)  +  dt sin t sin t 

∫ 

= cos ( a − b) ∫ 1 dt + sin ( a − b) ∫ cot ( t ) dt

= t cos ( a − b) + sin ( a − b) log |sin t | + C

= ( x + b) cos ( a − b) + sin ( a − b) log |sin ( x + b)| + C

Some Special Integrals (i)

∫ x2

1 dx x = tan − 1   + C  a a + a2

(ii)

∫ x2

 x − a dx 1 + C log  = 2 2 a −a  x + a

(iii)

∫ a2

(iv)

∫

(v)

∫

(vi)

∫

(vii)

∫

a2 − x2 dx =

a2 − x2 +

(viii)

∫

a2 + x2

a2 + x2

(ix)

∫

x2 − a2

a + x dx 1 + C log  = 2 2a −x a − x dx x = sin − 1   + C 2 2   a a −x dx a +x 2

2

dx x −a 2

2

= log | x +

x2 + a2 | + C

= log | x +

x2 − a2 | + C

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

x2 − a2

1 2 x a sin − 1   + C  a 2 1 + a2 log | x + a2 + x2 | + C 2 1 − a2 log | x + x2 − a2 | + C 2

Some Important Substitutions Expression

Substitution

a2 + x2

x = a tan θ or a cot θ

a −x

x = a sin θ or a cos θ

2

x −a a−x or a+x 2

x −α or β−x

2

x = a sec θ or a cosec θ

2

a+x a−x

x = a cos 2θ

( x − α ) ( x − β)

x = α cos 2 θ + β sin 2 θ