Evaluate $\displaystyle \int \limits_{0}^{\sqrt{2}} \sqrt{2 - x^2} \; dx$
Let $I = \displaystyle \int \limits_{0}^{\sqrt{2}} \sqrt{2 - x^2} \; dx$ $\;\;\; \cdots \; (1)$
Let $x = \sqrt{2} \sin \theta$ $\;\;\; \cdots \; (2)$
Differentiating equation $(2)$ gives
$dx = \sqrt{2} \cos \theta \; d\theta$ $\;\;\; \cdots \; (2a)$
From equation $(2)$, $\theta = \sin^{-1} \left(\dfrac{x}{\sqrt{2}}\right)$
When $x = 0$, $\theta = \sin^{-1} \left(\dfrac{0}{\sqrt{2}}\right) = 0$ $\;\;\; \cdots \; (2b)$
When $x = \sqrt{2}$, $\theta = \sin^{-1} \left(\dfrac{\sqrt{2}}{\sqrt{2}}\right) = \sin^{-1} \left(1\right) = \dfrac{\pi}{2}$ $\;\;\; \cdots \; (2c)$
$\therefore$ $\;$ In view of equations $(2)$, $(2a)$, $(2b)$ and $(2c)$, equation $(1)$ becomes
$\begin{aligned}
I & = \int \limits_{0}^{\frac{\pi}{2}} \sqrt{2 - 2 \sin^2 \theta} \times \sqrt{2} \cos \theta \; d\theta \\\\
& = \sqrt{2} \int \limits_{0}^{\frac{\pi}{2}} \sqrt{2 \left(1 - \sin^2 \theta\right)} \cos \theta \; d\theta \\\\
& = 2 \int \limits_{0}^{\frac{\pi}{2}} \cos \theta \times \cos \theta \; d\theta \\\\
& = 2 \int \limits_{0}^{\frac{\pi}{2}} \cos^2 \theta \; d\theta \\\\
& = 2 \int \limits_{0}^{\frac{\pi}{2}} \left(\dfrac{1 + \cos 2 \theta}{2}\right) \; d\theta \\\\
& = \int \limits_{0}^{\frac{\pi}{2}} d\theta + \int \limits_{0}^{\frac{\pi}{2}} \cos 2 \theta \; d\theta \\\\
& = \left[\theta\right]_{0}^{\pi / 2} + \dfrac{1}{2} \left[\sin 2 \theta\right]_{0}^{\pi / 2} \\\\
& = \dfrac{\pi}{2} - 0 + \dfrac{1}{2} \left(\sin \pi - \sin 0\right) \\\\
& = \dfrac{\pi}{2} + \dfrac{1}{2} \left(0 - 0\right) \\\\
& = \dfrac{\pi}{2}
\end{aligned}$