Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12

# 1 ∫ 0 Cos − 1 ( 1 − X 2 1 + X 2 ) D X - Mathematics

Sum

$\int\limits_0^1 \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) dx$

#### Solution

$\int_0^1 \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) d x$

$Let, x = \tan\theta, dx = se c^2 \theta d\theta$

$\text{When, }x \to 0 ; \theta \to 0$

$\text{and }x \to 1 ; \theta \to \frac{\pi}{4}$

Therefore, the integral becomes

$\int_0^\frac{\pi}{4} \cos^{- 1} \left( \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \right) se c^2 \theta d\theta$

$= \int_0^\frac{\pi}{4} \cos^{- 1} \left( \cos2\theta \right) se c^2 \theta d\theta$

$= 2 \int_0^\frac{\pi}{4} \theta se c^2 \theta d\theta$

$= 2 \left[ \theta tan\theta \right]_0^\frac{\pi}{4} - 2 \int_0^\frac{\pi}{4} \tan\theta d\theta$

$= 2 \left[ \theta \tan\ theta \right]_0^\frac{\pi}{4} + 2 \left[ \log\left( \cos\theta \right) \right]_0^\frac{\pi}{4}$

$= 2\left( \frac{\pi}{4} - 0 \right) + 2\left[ \log\frac{1}{\sqrt{2}} - 0 \right]$

$= \frac{\pi}{2} - \log2$

Concept: Definite Integrals Problems
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#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 20 Definite Integrals
Revision Exercise | Q 6 | Page 121