Question:
$\int_{0}^{1} \frac{d x}{\sqrt{1-x^{2}}}$
Solution:
Let $I=\int_{0}^{1} \frac{d x}{\sqrt{1-x^{2}}}$
$\int \frac{d x}{\sqrt{1-x^{2}}}=\sin ^{-1} x=\mathrm{F}(x)$
By second fundamental theorem of calculus, we obtain
$\begin{aligned} I &=\mathrm{F}(1)-\mathrm{F}(0) \\ &=\sin ^{-1}(1)-\sin ^{-1}(0) \\ &=\frac{\pi}{2}-0 \\ &=\frac{\pi}{2} \end{aligned}$