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

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

Answers (1)

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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