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

Sum

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

#### Solution

$I = \int_0^1 ( \cos^{- 1} x )^2 d x$

$\text{let }co s^{- 1} x = \theta$

$\Rightarrow x = \cos\theta$

$\Rightarrow dx = - \sin\theta d\theta$

$\text{when }x = 0, \theta = \frac{\pi}{2}\text{ and when }x = 1, \theta = 0$

$\text{Therefore, }I = \int_\frac{\pi}{2}^0 \theta^2 ( - \sin\theta) d \theta$

$I = - \int_\frac{\pi}{2}^0 \theta^2 (sin\theta) d \theta$

$I = \int_0^\frac{\pi}{2} \theta^2 (sin\theta) d \theta$

$I = \left[ \theta^2 ( - cos\theta) \right]_0^\frac{\pi}{2} - \int_0^\frac{\pi}{2} 2\theta \int_0^\frac{\pi}{2} \sin\theta d \theta$

$I = \left[ \theta^2 ( - \cos\theta) \right]_0^\frac{\pi}{2} - \int_0^\frac{\pi}{2} 2\theta( - \cos\theta) d \theta$

$= [ - \theta^2 \cos\theta ]_0^\frac{\pi}{2} + \int_0^\frac{\pi}{2} 2\theta(\cos\theta)d\theta$

$= [ - \theta^2 cos\theta ]_0^\frac{\pi}{2} + 2[\theta\sin\theta - \int_0^\frac{\pi}{2} \sin\theta d\theta]$

$= [ - \theta^2 cos\theta ]_0^\frac{\pi}{2} + 2[\theta sin\theta + \cos\theta ]_0^\frac{\pi}{2}$

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

$I = \pi - 2$

Concept: Definite Integrals Problems
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#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 20 Definite Integrals
Revision Exercise | Q 25 | Page 121