Question

Evaluate the definite integral ${\int }_0^1\frac{dx}{\sqrt{1-x^2}}$

Answer 1

Let $I={\int }_0^1\frac{dx}{\sqrt{1-x^2}}$
$\Rightarrow \int \frac{dx}{\sqrt{1-x^2}}={\sin }^{-1}x=F\left(x\right)$
By second fundamental theorem of calculus, we obtain
$I=F\left(1\right)-F\left(0\right)$
$={\sin }^{-1}\left(1\right)-{\sin }^{-1}\left(0\right)$
$=\frac{\pi }{2}-0=\frac{\pi }{2}$