Question

Evaluate:

${\int }^{}\frac{-1}{\sqrt{1-x^2}}dx$

Answer 1

Put x = sint $\Rightarrow \frac{dx}{dt}=\cos tdt$
$\therefore I={\int }^{}\frac{-\cos tdt}{\sqrt{1-{\sin }^{2}t}}={\int }^{}\frac{-\cos tdt}{{\cos }^{2}t}$
$=-t={\sin }^{-1}x$
${\cos }^{-1}x+{\sin }^{-1}x=\frac{\pi }{2}$
$\Rightarrow {\cos }^{-1}-\frac{\pi }{2}=-{\sin }^{-1}x$
$\therefore t={\cos }^{-1}x-\frac{\pi }{2}$
Here, $C=-\frac{\pi }{2}$