Question
Solve: $\int \cos ^{-1}(\sin x) d x$

Answer

$
\begin{aligned}
I & =\int \cos ^{-1}(\sin x) d x \\
& =\int \cos ^{-1}\left(\cos \left(\frac{\pi}{2}-x\right)\right) d x \\
& =\int\left(\frac{\pi}{2}-x\right) d x=\frac{\pi}{2} \int d x-\int x d x \\
& =\frac{\pi}{2} x-\frac{x^2}{2}+C \quad \text { Ans. }
\end{aligned}
$

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