Question

Find the integral of the function ${\sin }^{-1}\left(\cos x\right)$

Answer 1

Let $\cos x=t$
Then, $\sin x=\sqrt{1-t^2}$
$\Rightarrow (-\sin x)dx=dt$
$\Rightarrow dx=\frac{-dt}{\sin x}$
$\Rightarrow dx=\frac{-dt}{\sqrt{1-t^2}}$
$\therefore \int {\sin }^{-1}\left(\cos x\right)dx=\int {\sin }^{-1}t\left(\frac{-dt}{\sqrt{1-t^2}}\right)$
$=-\int \frac{{\sin }^{-1}t}{\sqrt{1-t^2}}dt$
Let ${\sin }^{-1}t=u$
$\Rightarrow \frac{1}{\sqrt{1-t^2}}dt=du$
$\therefore \int {\sin }^{-1}\left(\cos x\right)dx=\int 4du$
$=-\frac{u^2}{2}+C$
$=\frac{-({\sin }^{1}t{)}^2}{2}+C$
$=\frac{-\left[{\sin }^{-1}\right(\cos x){]}^2}{2}+C$ .......(1)
It is known that,
${\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}$
$\therefore {\sin }^{-1}\left(\cos x\right)=\frac{\pi }{2}-{\cos }^{-1}\cos x=(\frac{\pi }{2}-x)$
Substituting in equation (1), we obtain
$\int {\sin }^{-1}\left(\cos x\right)dx=\frac{-{[\frac{\pi }{2}-x]}^2}{2}+C$
$=-\frac{1}{2}(\frac{{\pi }^2}{2}+x^2-\pi x)+C$
$=-\frac{{\pi }^2}{8}-\frac{x^2}{2}+\frac{1}{2}\pi x+C$
$=\frac{\pi x}{2}-\frac{x^2}{2}+(C-\frac{{\pi }^2}{8})$
$=\frac{\pi x}{2}-\frac{x^2}{2}+C_1$