Evaluate ∫ e 2 z e 4 z + 6 d z

Let I = ∫ e 2 z e 4 z + 6 d z = ∫ e 2 z e 2 z 2 + 6 2 d z Putting e 2 z = t ⇒ 2 e 2 z d z = d t So, I = 1 2 ∫ t 2 + ( 6 ) 2 d t = 1 2 t t 2 + 6 2 + 6 2 ln t + t 2 + 6 + c = e 2 z e 4 z + 6 4 + 3 ln e 2 z + e 4 z + 6 2 + c

Evaluate ∫ sec 2 ⁡ θ sec 2 ⁡ θ + 3 d θ

Let I = ∫ sec 2 ⁡ θ sec 2 ⁡ θ + 3 d θ = ∫ sec 2 ⁡ θ tan 2 ⁡ θ + 4 d θ ∵ sec 2 ⁡ x = tan 2 ⁡ x + 1 Putting u = tan θ ⇒ d u = sec 2 θ d θ So, I = ∫ u 2 + 2 2 d u = u 2 u 2 + 4 + 4 2 ln u + u 2 + 4 + c = u 2 u 2 + 4 + 2 ln u + u 2 + 4 + c = tan θ 2 tan 2 θ + 4 + 2 ln tan θ + tan 2 θ + 4 + c = tan θ 2 sec 2 θ + 3 + 2 ln tan θ + sec 2 θ + 3 + c ∵ sec 2 ⁡ x = tan 2 ⁡ x + 1

Evaluate ∫ 2 x 2 + 1 e x 2 d x

Let I = ∫ 2 x 2 + 1 e x 2 d x = ∫ 2 x 2 e x 2 + e x 2 d x = 2 ∫ x 2 e x 2 d x + ∫ e x 2 d x Consider, I 1 = ∫ e x 2 · 1 d x Applying Integration By Parts by taking e x 2 as first function and 1 as second function. = e x 2 ∫ 1 d x - ∫ d e x 2 d x ∫ 1 d x d x = e x 2 · x - ∫ 2 x e x 2 · x d x So, I = 2 ∫ x 2 e x 2 d x + e x 2 ⋅ x − ∫ 2 x 2 e x 2 d x ⇒ I = x e x 2 + c

Evaluate ∫ x 2 − 8 d x

Let I = ∫ x 2 − 8 d x = ∫ x 2 − 8 2 d x = x x 2 - 8 2 - 8 ln x + x 2 - 8 2 + c = x x 2 - 8 2 - 4 ln x + x 2 - 8 + c

Evaluate ∫ 3 x 2 − 2 d x

Let I = ∫ 3 x 2 − 2 d x = ∫ 3 x 2 − 2 3 d x = 3 ∫ x 2 − 2 3 2 d x = 3 x x 2 - 2 3 2 - 1 2 × 2 3 ln x + x 2 - 2 3 + c = 3 x 3 x 2 - 2 3 × 2 - 1 3 ln x + 3 x 2 - 2 3 + c = x 3 x 2 - 2 2 - 1 3 ln x + 3 x 2 - 2 3 + c

Evaluate ∫ a z a 2 z − 4 d z

Let I = ∫ a z a 2 z − 4 d z = ∫ a z a z 2 − 2 2 d z Putting a z = t ⇒ a z ln a d z = d t So, I = 1 ln a ∫ t 2 - 2 2 d t = 1 ln a t t 2 - 4 2 - 4 2 ln t + t 2 - 4 + c = 1 ln a a z a 2 z - 4 2 - 2 ln a z + a 2 z - 4 + c

Evaluate ∫ sec ⁡ θ tan ⁡ θ tan 2 ⁡ θ − 3 d θ

Let I = ∫ sec ⁡ θ tan ⁡ θ tan 2 ⁡ θ − 3 d θ = ∫ sec ⁡ θ tan ⁡ θ sec 2 ⁡ θ − 4 d θ ∵ 1 + tan 2 x = sec 2 x Putting sec θ = t ⇒ sec θ tan θ d θ = d t So, I = ∫ t 2 - 2 2 d t = t t 2 - 4 2 - 4 2 ln t + t 2 - 4 + c = sec ⁡ θ sec 2 ⁡ θ - 4 2 - 2 ln sec ⁡ θ + sec 2 ⁡ θ - 4 + c = sec ⁡ θ tan 2 ⁡ θ - 3 2 - 2 ln sec ⁡ θ + tan 2 ⁡ θ - 3 + c ∵ 1 + tan 2 ⁡ x = sec 2 ⁡ x

Evaluate ∫ sin ⁡ x ln ⁡ ( cosec ⁡ x − cot ⁡ x ) d x

Let I = ∫ sin ⁡ x ln ⁡ ( cosec ⁡ x − cot ⁡ x ) d x Applying Integration By parts by taking ln ⁡ ( cosec ⁡ x − cot ⁡ x ) as first function and sin x as second function. So, I = ln ⁡ ( cosec ⁡ x − cot ⁡ x ) ∫ sin x d x - ∫ d ln ⁡ ( cosec ⁡ x − cot ⁡ x ) d x ∫ sin x d x d x = - cos x · ln ⁡ ( cosec ⁡ x − cot ⁡ x ) + ∫ - cosec x cot x + cosec 2 x cosec ⁡ x − cot ⁡ x · cos x d x = - cos x · ln ⁡ ( cosec ⁡ x − cot ⁡ x ) + ∫ cosec x cosec x - cot x cosec ⁡ x − cot ⁡ x · cos x d x = - cos x · ln ⁡ ( cosec ⁡ x − cot ⁡ x ) + ∫ cot x d x = - cos x · ln ⁡ ( cosec ⁡ x − cot ⁡ x ) + ln sin x + c

E M B I B E

Elements of Mathematics Class 12>Chapter 9 - Integration>EXERCISE-9(e)

Odisha Board Solutions for Chapter: Integration, Exercise 5: EXERCISE-9(e)

Author:Odisha Board

Odisha Board Mathematics Solutions for Exercise - Odisha Board Solutions for Chapter: Integration, Exercise 5: EXERCISE-9(e)

Attempt the practice questions on Chapter 9: Integration, Exercise 5: EXERCISE-9(e) with hints and solutions to strengthen your understanding. Elements of Mathematics Class 12 solutions are prepared by Experienced Embibe Experts.