$\sin ^{-1}(\cos x)$
$\sin ^{-1}(\cos x)$
Let $\cos x=t$
Then, $\sin x=\sqrt{1-t^{2}}$
$\Rightarrow(-\sin x) d x=d t$
$d x=\frac{-d t}{\sin x}$
$d x=\frac{-d t}{\sqrt{1-t^{2}}}$
$\therefore \int \sin ^{-1}(\cos x) d x=\int \sin ^{-1} t\left(\frac{-d t}{\sqrt{1-t^{2}}}\right)$
$=-\int \frac{\sin ^{-1} t}{\sqrt{1-t^{2}}} d t$
Let $\sin ^{-1} t=u$
$\Rightarrow \frac{1}{\sqrt{1-r^{2}}} d t=d u$
$\therefore \int \sin ^{-1}(\cos x) d x=\int 4 d u$
$=-\frac{u^{2}}{2}+C$
$=\frac{-\left(\sin ^{1} t\right)^{2}}{2}+C$
$=\frac{-\left[\sin ^{-1}(\cos x)\right]^{2}}{2}+C$ ....(1)
It is known that,
$\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}$
$\therefore \sin ^{-1}(\cos x)=\frac{\pi}{2}-\cos ^{-1}(\cos x)=\left(\frac{\pi}{2}-x\right)$
Substituting in equation (1), we obtain
$\int \sin ^{-1}(\cos x) d x=\frac{-\left[\frac{\pi}{2}-x\right]^{2}}{2}+C$
$=-\frac{1}{2}\left(\frac{\pi^{2}}{2}+x^{2}-\pi x\right)+C$
$=-\frac{\pi^{2}}{8}-\frac{x^{2}}{2}+\frac{1}{2} \pi x+C$
$=\frac{\pi x}{2}-\frac{x^{2}}{2}+\left(C-\frac{\pi^{2}}{8}\right)$
$=\frac{\pi x}{2}-\frac{x^{2}}{2}+C_{1}$
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