Question

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

Answer 1

${\int }_0^1\frac{1}{\sqrt{1-x^2}}dx=[si{n}^{-1}x{]}_0^1(\because \int \frac{dx}{\sqrt{1-x^2}}=si{n}^{-1}x)=si{n}^{-1}1-si{n}^{-1}0=\frac{\pi }{2}-0=\frac{\pi }{2}$