Question

Solve :
$\int \frac{dx}{\sqrt{1-x^2}}={\sin }^{-1}x+c$

Answer 1

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

Put $x=\sin \theta$, we get $dx=\cos \theta d\theta$

So, $I=\int \frac{d\theta \times \cos \theta }{\sqrt{1-{\sin }^2\theta }}=\int \frac{d\theta \times \cos \theta }{\cos \theta }=\int d\theta =\theta +c$

$I=\theta +c$

$x=\sin \theta ⟹\theta ={\sin }^{-1}x$

So, $I={\sin }^{-1}x+c$