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Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise Very Short Answers Question 60

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Answer: \frac{\pi x}{2}-\frac{x^{2}}2{}+c

Hints: You must know about the rules of integration of trigonometric and inverse trigonometric functions.

Given: \int \cos ^{-1}\left ( \sin x \right )dx


\int \cos ^{-1}\left ( \sin x \right )dx

Let \cos ^{-1}\left ( \sin x \right )=\theta

\begin{aligned} &\sin x=\cos \theta \\ &\sin x=\sin \left(\frac{\pi}{2}-\theta\right) \\ &x=\left(\frac{\pi}{2}-\theta\right) \\ &\theta=\frac{\pi}{2}-x \\ &\therefore \int \cos ^{-1}(\sin x)dx=\int \left ( \frac{\pi}{2}-x \right )dx \\ &=\int \frac{\pi}{2} d x-\int x dx \\ \end{aligned}

=\frac{\pi x}{2}-\frac{x^{2}}2{}+c

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