Question

Evaluate :

${{\int }_0}^1\frac{dx}{\sqrt{1-x^2}}$

Answer 1

$I={{\int }_0}^1\frac{dx}{\sqrt{1-x^2}}$

$\int \frac{dx}{\sqrt{1-x^2}}={\sin }^{-1}x=F\left(x\right)$

By second fundamental theorem of calculus,

$I=F\left(1\right)-F\left(0\right)$

$I={\sin }^{-1}\left(1\right)-{\sin }^{-1}\left(0\right)$

$I=\frac{\pi }{2}-0=\frac{\pi }{2}$