Question

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

Answer 1

Consider the given integral.

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

We know that

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

Therefore,

$I={\left[{\sin }^{-1}\left(x\right)\right]}_0^1$

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

$I={\sin }^{-1}\left(\sin \frac{\pi }{2}\right)-{\sin }^{-1}\left(\sin 0\right)$

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

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

Hence, this is the answer