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Need solution for RD Sharma Maths Class 12 Chapter 19 Definite Integrals Excercise Revision Exercise Question 25

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Answer: $\pi-2$

Given: $\int_{0}^{1}\left(\cos ^{-1} x\right)^{2}$

Hint: Using substitution method and formula of $\int uv dx$

Solution: $\int_{0}^{1}\left(\cos ^{-1} x\right)^{2}$

Let

$\begin{aligned} &\cos ^{-1} x=\theta \Rightarrow x=\cos \theta \\ & \end{aligned}$ (Differentiate w.r.t to x)

$d x=-\sin \theta d \theta \\$

$=\int_{0}^{1} \theta^{2}(-\sin \theta d \theta)$

$\begin{aligned} &=-\int_{0}^{1} \theta^{2} \sin \theta d \theta \\ & \end{aligned}$

$=-\left[\theta^{2}(\cos \theta)\right]_{0}^{1}+\int_{0}^{1} 2 \theta(-\cos \theta) d \theta \\$

$=\theta^{2}(\cos \theta)_{0}^{1}-2\left[(\theta \sin \theta)_{0}^{1}-\int_{0}^{1} \sin \theta d \theta\right]$

$\begin{aligned} &=(\theta^{2}\cos \theta)_{0}^{1}-2(\theta \sin \theta)_{0}^{1}+2(-\cos \theta)_{0}^{1} \\ \end{aligned}$

$=\left[\left(\cos ^{-1} x\right)^{2} x\right]_{0}^{1}-2\left[\cos ^{-1} x \sin \left(\cos ^{-1} x\right)\right]_{0}^{1}-2(x)_{0}^{1} \\$

$= 0-2\left(0-\frac{\pi}{2}\right)-2=\frac{2 \pi}{2}-2 \\$

$= \pi-2$

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