Evaluate ∫ 0 π d x 6 - cos x

Let I = ∫ 0 π d x 6 - cos x Put tan x 2 = t ⇒ sec 2 x 2 . 1 2 d x = d t ⇒ d x = 2 d t 1 + t 2 cos x = 1 - t 2 1 + t 2 When x = 0 ⇒ t = 0 and when x = π 2 ⇒ t = 1 ∴ I = ∫ 0 1 2 d t 1 + t 2 6 - 1 - t 2 1 + t 2 = ∫ 0 1 2 d t 6 + 6 t 2 - 1 + t 2 = 2 ∫ 0 1 d t 5 + 7 t 2 = 2 7 ∫ 0 1 d t t 2 + 5 7 2 = 2 7 × 7 5 tan - 1 7 t 5 0 1 = 2 7 × 7 5 tan - 1 7 5 - tan - 1 0 = 2 35 tan - 1 7 5

Evaluate ∫ 0 1 1 - x 2 x 4 + x 2 + 1 d x

Let I = ∫ 0 1 1 - x 2 d x x 4 + x 2 + 1 Divide numerator and denominator by x 2 we have I = ∫ 0 1 1 x 2 - 1 x 2 + 1 + 1 x 2 Put x + 1 x = t ⇒ 1 - 1 x 2 d x = d t On squaring both sides x + 1 x 2 = t 2 ⇒ x 2 + 1 x 2 + 2 = t 2 ⇒ x 2 + 1 x 2 = t 2 - 2 When x = 0 ⇒ t → ∞ When x = 1 ⇒ t = 2 = ∫ ∞ 2 d t t 2 - 1 = ∫ ∞ 2 d t 1 2 - t 2 = 1 2 × 1 log 1 + t 1 - t ∞ 2 = 1 2 log 1 + 2 1 - 2 - limlog t 1 + 1 t t 1 t - 1 t → ∞ = 1 2 log 3 - log 1 = 1 2 log 3

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Understanding ISC Mathematics Class 12 Volume 2>Chapter 8 - Integrals>EXERCISE

M L Aggarwal Solutions for Chapter: Integrals, Exercise 17: EXERCISE

Author:M L Aggarwal

M L Aggarwal Mathematics Solutions for Exercise - M L Aggarwal Solutions for Chapter: Integrals, Exercise 17: EXERCISE

Attempt the free practice questions on Chapter 8: Integrals, Exercise 17: EXERCISE with hints and solutions to strengthen your understanding. Understanding ISC Mathematics Class 12 Volume 2 solutions are prepared by Experienced Embibe Experts.