Question

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

Answer 1

We have,

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

Let

$t={\cos }^{-1}x$

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

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

Therefore,

$I=-\int tdt$

$I=-\frac{t^2}{2}+C$

Put $t={\cos }^{-1}x$

$I=-\frac{({\cos }^{-1}x{)}^2}{2}+C$

Hence, this is the answer.