# Prove

Question:

$\sin ^{-1}(\cos x)$

Solution:

$\sin ^{-1}(\cos x)$

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

$\int \sin ^{-1}(\cos x) d x=\frac{-\left[\frac{\pi}{2}-x\right]^{2}}{2}+C$

$=-\frac{1}{2}\left(\frac{\pi^{2}}{2}+x^{2}-\pi x\right)+C$

$=-\frac{\pi^{2}}{8}-\frac{x^{2}}{2}+\frac{1}{2} \pi x+C$

$=\frac{\pi x}{2}-\frac{x^{2}}{2}+\left(C-\frac{\pi^{2}}{8}\right)$

$=\frac{\pi x}{2}-\frac{x^{2}}{2}+C_{1}$