Evaluate ∫ ( x - 4 ) e x ( x - 2 ) 3 d x .

I = ∫ ( x - 4 ) e x ( x - 2 ) 3 d x . ⇒ I = ∫ ( x - 2 - 2 ) e x ( x - 2 ) 3 d x . ⇒ I = ∫ ( x - 2 ) ( x - 2 ) 3 - 2 x - 2 3 e x d x . ⇒ I = ∫ 1 ( x - 2 ) 2 - 2 x - 2 3 e x d x Let f ( x ) = 1 x - 2 2 . Differentiating f ( x ) , with respect to x . d ( f ( x ) d x = d 1 x - 2 2 d x ⇒ f ' ( x ) = - 2 x - 2 3 ⇒ I = ∫ f ( x ) + f ' ( x ) e x d x ⇒ I = f ( x ) . e x + c ⇒ I = 1 x - 2 2 . e x + c , where c , is a constant.

Evaluate ∫ x 2 + x + 1 ( x + 2 ) x 2 + 1 d x .

I = ∫ x 2 + x + 1 ( x + 2 ) x 2 + 1 d x . As we can see that the integral in the form of ∫ Q L . Q 1 d y , then the Q , can be substituted with A L + B x + C Q 1 . . . ˙ x 2 + x + 1 ( x + 2 ) x 2 + 1 = A ( x + 2 ) + B x + C x 2 + 1 ⇒ . . ˙ x 2 + x + 1 ( x + 2 ) x 2 + 1 = A ( x 2 + 1 ) + B x + C ( x + 2 ) ( x + 2 ) ( x 2 + 1 ) On comparing the coefficients on both side of the equation. ⇒ A + B = 1 & 2 B + C = 1 & A + 2 C = 1 ⇒ A = 3 5 , B = 2 5 & C = 1 5 . . . ˙ I = ∫ 3 5 x + 2 + 2 5 x + 1 5 x 2 + 1 d x ⇒ I = 3 5 ∫ d x x + 2 + 2 5 ∫ x d x x 2 + 1 + 1 5 ∫ d x x 2 + 1 . ⇒ I = 3 5 ln x + 2 + c 1 + 2 5 ∫ x d x x 2 + 1 + 1 5 tan - 1 x 1 + c 2 , using d y y = ln y + c & ∫ d x x 2 + a 2 = 1 a tan - 1 x a + c , where c , c 1 , c 2 , are integration constant. Let t = x 2 + 1 Differentiate t , with respect to x . d t d x = d ( x 2 + 1 ) d x d t d x = 2 x , using d ( y n ) d y = n y n - 1 & d ( C ) d y = 0 , where C , is a constant. d t = 2 x d x . ⇒ I = 3 5 ln x + 2 + c 1 + 1 5 ∫ d t t + 1 5 tan - 1 x + c 2 ⇒ I = 3 5 ln x + 2 + c 1 + 1 5 ln t + c 3 + 1 5 tan - 1 x + c 2 . ⇒ I = 3 5 ln x + 2 + 1 5 ln x 2 + 1 + 1 5 tan - 1 x + C , where C , is an integration constant.

Evaluate ∫ tan x + tan 3 x 1 + tan 3 x d x .

I = ∫ tan x + tan 3 x 1 + tan 3 x d x . ⇒ I = ∫ tan x ( 1 + tan 2 x ) 1 + tan 3 x d x . ⇒ I = ∫ tan x × sec 2 x 1 + tan 3 x d x . Let t = tan x . Differentiating t , with respect to x . ⇒ d t d x = d ( tan x ) d x . ⇒ d t d x = sec 2 x ⇒ d t = sec 2 x d x ⇒ I = ∫ t d t 1 + t 3 Adding and subtracting in the numerator by t 2 . ⇒ I = ∫ t 2 - t 2 + t 1 + t 3 d t . ⇒ I = ∫ t 2 + t 1 + t 3 d t - ∫ t 2 1 + t 3 d t . ⇒ I = ∫ t ( 1 + t ) ( 1 + t ) ( 1 + t 2 - t ) d t - ∫ t 2 1 + t 3 d t , using ( a 3 + b 3 ) = ( a + b ) ( a 2 + b 2 - a b ) . ⇒ I = ∫ t ( 1 + t 2 - t ) d t - ∫ t 2 1 + t 3 d t . Let 1 + t 3 = u . Differentiate u , with respect to t d u d t = d ( 1 + t 3 ) d t . d u 3 = t 2 d t . ⇒ I = ∫ t ( 1 + t 2 - t ) d t - 1 3 ∫ d u u . ⇒ I = ∫ t ( 1 + t 2 - t ) d t - 1 3 ln u + c 1 , using d x x = ln x + C ' , where C ' , is an integration constant. ⇒ I = ∫ t ( 1 + t 2 - t ) d t - 1 3 ln 1 + tan 3 x + c 1 . Now dividing and multiplying the numerator of first integration by 2 . ⇒ I = 1 2 ∫ 2 t ( 1 + t 2 - t ) d t - ln 1 + tan 3 x + c 1 . ⇒ I = 1 2 ∫ 2 t - 1 + 1 ( 1 + t 2 - t ) d t - ln 1 + tan 3 x + c 1 . ⇒ ⇒ I = 1 2 ∫ 2 t - 1 2 ( 1 + t 2 - t ) d t + 1 2 ∫ 1 ( 1 + t 2 - t ) d t - 1 3 ln 1 + tan 3 x + c 1 . Let 1 + t 2 - t = v , from the first integration of I . Differentiating v , with respect to t . ⇒ d v d t = d ( 1 + t 2 - t ) d t ⇒ d v = 2 t - 1 d t . . . ˙ ⇒ I = 1 2 ∫ d v v + 1 2 ∫ 1 ( 1 + t 2 - t ) d t - 1 3 ln 1 + tan 3 x + c 1 . . . ˙ ⇒ I = 1 2 ln v + c 2 + 1 2 ∫ 1 t - 1 2 2 + 3 2 2 d t - 1 3 ln 1 + tan 3 x + c 1 , using completing square method in the denominator. ⇒ I = 1 2 ln t 2 - t + 1 + c 2 + 1 2 × 1 3 2 tan - 1 t - 1 2 3 2 + c 3 - 1 3 ln 1 + tan 3 x + c 1 , using 1 x 2 + a 2 = 1 a tan - 1 x a + C , where C , is an integration constant. Here c 1 , c 2 & c 3 , are the integration constant. ⇒ I = 1 2 ln t 2 - t + 1 + 3 tan - 1 2 t - 1 2 3 - 1 3 ln 1 + tan 3 x + c . ⇒ I = 1 2 ln t 2 - t + 1 + 3 tan - 1 2 t - 1 3 - 1 3 ln 1 + tan 3 x + c . ⇒ I = 1 2 ln tan 2 x - tan x + 1 + 1 3 tan - 1 2 tan x - 1 3 - 1 3 ln 1 + tan 3 x + c .

Evaluate ∫ sin 2 x ( 1 - cos 2 x ) ( 2 - cos 2 x ) d x .

I = ∫ sin 2 x ( 1 - cos 2 x ) ( 2 - cos 2 x ) d x . Let t = cos 2 x Differentiating t , with respect to x . ⇒ d t d x = d ( cos 2 x ) d x . ⇒ d t d x = - 2 sin ( 2 x ) , using d ( cos y ) d y = - sin y & d ( 2 y ) d y = 2 . ⇒ - d t 2 = sin ( 2 x ) d x Therefore I = - 1 2 ∫ d t ( 1 - t ) ( 2 - t ) . ⇒ I = - 1 2 ∫ ( 2 - t ) - ( 1 - t ) ( 1 - t ) ( 2 - t ) d t . ⇒ I = - 1 2 ∫ ( 2 - t ) ( 1 - t ) ( 2 - t ) d t - ∫ ( 1 - t ) ( 1 - t ) ( 2 - t ) d t . ⇒ I = - 1 2 ∫ 1 ( 1 - t ) d t - ∫ 1 ( 2 - t ) d t . ⇒ I = - 1 2 - log | 1 - t | - ( - log | 2 - t | ) + c , using ∫ 1 a x + b d x = 1 a log a x + b + C , where C , is an integration constant. ⇒ I = - 1 2 log | 2 - t | - log | 1 - t | + c , where c , is an integration constant. ⇒ I = 1 2 log | 1 - cos 2 x | - log | 2 - cos 2 x | + c Using log a - log b = log a b . I = 1 2 log 1 - cos 2 x 2 - cos 2 x + c .

Evaluate ∫ 2 x + 1 2 x 2 + 4 x - 3 d x .

I = ∫ 2 x + 1 2 x 2 + 4 x - 3 d x . As we can see that the given integral is in the form of ∫ P Q d x , where P , is a linear equation and Q , is a quadratic equation. So we can write P = A d ( Q ) d x + B . Substitute 2 x + 1 = A d ( 2 x 2 + 4 x + 3 ) d x + B . ⇒ 2 x + 1 = A ( 4 x + 4 ) + B . On comparing the coefficients on both sides we get. ⇒ 4 A = 2 & 4 A + B = 1 . A = - 1 2 & B = - 1 . ⇒ I = ∫ 1 2 ( 4 x + 4 ) - 1 2 x 2 + 4 x - 3 d x Substitute 2 x + 1 = A d ( 2 x 2 + 4 x - 3 ) d x + B . ⇒ 2 x + 1 = 4 x + 4 A + B . ⇒ 2 x + 1 = 4 A x + 4 A + B . On comparing both side we get. ⇒ 2 = 4 A & 4 A + B = 1 . ⇒ A = 1 2 & B = - 1 . ⇒ I = ∫ - 1 2 ( 4 x + 4 ) - 1 2 x 2 + 4 x - 3 d x . ⇒ I = ∫ - 1 2 ( 4 x + 4 ) 2 x 2 + 4 x - 3 d x - ∫ 1 2 x 2 + 4 x - 3 d x . Let I 1 = ∫ - 1 2 ( 4 x + 4 ) 2 x 2 + 4 x - 3 d x & I 2 = ∫ 1 2 x 2 + 4 x - 3 d x . ⇒ I 1 = ∫ - 1 2 ( 4 x + 4 ) 2 x 2 + 4 x - 3 d x . Let t = 2 x 2 + 4 x - 3 Differentiate t , with respect to x . ⇒ d t d x = d ( 2 x 2 + 4 x - 3 ) d x . ⇒ d t d x = 4 x + 4 ⇒ d t = 4 x + 4 d x . . . ˙ I 1 = 1 2 ∫ d t t . Using ∫ d x x = ln x + c ' , where c ' , is an integration constant. ⇒ I 1 = 1 2 ln t + c 1 , where c 1 , is an integration constant. ⇒ I 1 = 1 2 ln 2 x 2 + 4 x - 3 + c 1 Now I 2 = ∫ d x 2 x 2 + 4 x - 3 . ⇒ I 2 = 1 2 ∫ d x x 2 + 2 x - 3 2 . Adding and subtracting 1 , in the denominator. I 2 = 1 2 ∫ d x x 2 + 2 x - 3 2 + 1 - 1 . Using completing square method. I 2 = 1 2 ∫ d x ( x + 1 ) 2 - 5 2 . I 2 = 1 2 ∫ d x ( x + 1 ) 2 - 5 2 2 . As we know d x x 2 - a 2 = 1 2 a ln x - a x + a + c , where c , is an integration constant. ⇒ I 2 = 1 2 × 1 2 × 5 2 ln x + 1 - 5 2 x + 1 + 5 2 + c 2 , where c 2 , is an integration constant. ⇒ I 2 = 1 2 2 × 5 ln x + 1 - 5 2 x + 1 + 5 2 + c 2 . Now putting I 1 & I 2 , in the I . I = 1 2 ln 2 x 2 + 4 x - 3 - 1 2 2 × 5 ln x + 1 - 5 2 x + 1 + 5 2 + C , where C , is an integration constant

Evaluate ∫ ( 2 x - 5 ) x 2 - 4 x + 3 d x .

I = ∫ ( 2 x - 5 ) x 2 - 4 x + 3 d x . AS we can see that the integral is written in the form of ∫ L Q d x . , where L , is a linear function and Q , is a quadratic function. Using L = a Q ' + b , where Q ' , is the derivative of the quadratic function. Substitute 2 x - 5 = a d ( x 2 - 4 x + 3 ) d x + b . ⇒ 2 x - 5 = a ( 2 x - 4 ) + b . ⇒ 2 x - 5 = 2 a x - 4 a + b . On comparing the coefficients on both the side we get. ⇒ 2 a = 2 & b - 4 a = - 5 . ⇒ a = 1 & b = - 1 . ⇒ I = ∫ 1 × ( 2 x - 4 ) - 1 x 2 - 4 x + 3 . ⇒ I = ∫ 2 x - 4 x 2 - 4 x + 3 d x - 1 ∫ x 2 - 4 x + 3 d x . Let t = x 2 - 4 x + 3 . Differentiate t , with respect to x d t d x = d ( x 2 - 4 x + 3 ) d x . d t = 2 x - 4 d x . ⇒ I = ∫ t d t - ∫ x 2 - 4 x + ( 2 ) 2 - 1 d x . ⇒ I = ∫ t 1 2 d t - ∫ ( x + 2 ) 2 - 1 2 d x , using completing square method. ⇒ I = t 1 2 + 1 1 2 + 1 - x + 2 2 ( x + 2 ) 2 - 1 2 - ( x + 2 ) 2 2 ln ( x + 2 ) + ( x + 2 ) 2 - 1 2 + C , using ∫ y 1 2 d y = y 1 2 + 1 1 2 + 1 + c , and ∫ x 2 - a 2 d x = x 2 x 2 - a 2 - a 2 2 ln ( x + x 2 - a 2 ) . ⇒ I = 2 x 2 - 4 x + 3 3 3 - x + 2 2 x 2 - 4 x + 3 - ( x + 2 ) 2 2 ln ( x + 2 ) + x 2 - 4 x + 3 + C , where C , is a constant.

Evaluate ∫ x x + x 2 d x .

I = ∫ x x + x 2 d x . Using partial fraction. x = A d ( x + x 2 ) d x + B . ⇒ x = A ( 1 + 2 x ) + B ., using d ( x n ) d x = n x n - 1 ⇒ x = A + 2 A x + B . On comparing the equation on both side we can write. 2 A x = x & A + B = 0 ⇒ A = 1 2 & B = - 1 2 So we can write x = 1 2 ( 1 + 2 x ) - 1 2 . ⇒ I = ∫ 1 2 ( 1 + 2 x ) x + x 2 - 1 2 x + x 2 d x . ⇒ I = 1 2 ∫ ( 1 + 2 x ) x + x 2 d x - 1 2 ∫ x + x 2 d x . Let t = x + x 2 . Differentiating t , with respect to x . ⇒ d ( t ) d x = d ( x ) d x + d ( x 2 ) d x d t = 1 + 2 x d x , using d ( x n ) d x = n x n - 1 . ⇒ I = 1 2 ∫ t d t - 1 2 ∫ 2 × x 2 + x 2 - 1 4 + 1 4 d x . Using the identity ( a + b ) 2 = a 2 + b 2 + 2 a b . ⇒ I = 1 2 t 1 2 + 1 1 2 + 1 - 1 2 ∫ x + 1 4 2 - 1 4 d x , using x n d x = x n + 1 n + 1 + c . ⇒ I = t 3 2 3 - 1 2 x + 1 4 2 x + 1 4 2 - 1 4 - 1 8 ln ( x + 1 4 ) + x + 1 4 2 - 1 4 + c , using x 2 - a 2 = x 2 x 2 - a 2 - a 2 2 ln ( x + x 2 - a 2 ) . ⇒ I = x + x 2 3 3 - 1 2 x + 1 4 2 x + x 2 - 1 8 ln ( x + 1 4 ) + x + x 2 + c , where c , is an integration constant.

Evaluate ∫ sin 2 x ( a + b cos x ) 2 d x .

⇒ I = ∫ sin 2 x a + b cos x 2 . Using sin 2 A = 2 sin A cos A , we can write. I = ∫ 2 sin x cos x a + b cos 2 . Let t = a + b cos x . Differentiating t , with respect to x . ⇒ d t d x = d ( a ) d x + d ( b cos x ) d x . d t d x = - b sin x ⇒ d t b = - sin x . Now, I = 2 ∫ t - a b t 2 d t . ⇒ I = - 2 b ∫ t t 2 d t - ∫ a t 2 d t I = - 2 b ∫ 1 t d t - ∫ a × t 2 d t . I = - 2 b ln t - a t - 2 + 1 - 2 + 1 + C , using ∫ 1 x d x = ln x + c & ∫ x n d x = x n + 1 n + 1 + c , where C & c , are integration constant. I = - 2 b ln a + b cos x + a ( a + b cos x ) + C .

E M B I B E

Gamma Question Bank for Engineering Mathematics>Class XII>Chapter 26 - Indefinite Integrals>Assignment Set-1

Embibe Experts Solutions for Exercise 5: Assignment Set-1

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Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Exercise 5: Assignment Set-1

Attempt the free practice questions from Exercise 5: Assignment Set-1 with hints and solutions to strengthen your understanding. Gamma Question Bank for Engineering Mathematics solutions are prepared by Experienced Embibe Experts.