Find the area enclosed by the circle $x^2 + y^2 = a^2$
$x^2 + y^2 = a^2$ $\;\;\; \cdots \; (1)$
is a circle with center at $\left(0,0\right)$ and radius $= a$ units.
The circle cuts the positive X and Y axis at the points $\left(a,0\right)$ and $\left(0,a\right)$ respectively.
Required area $= A = 4 \left|I\right|$ where
$I = \displaystyle \int \limits_{0}^{a} y \; dx$ $\;\;\; \cdots \; (2)$
We have from equation $(1)$, $y = \sqrt{a^2 - x^2}$ $\;\;\; \cdots \; (3)$
$\therefore$ $\;$ In view of equation $(3)$, equation $(2)$ becomes
$I = \displaystyle \int \limits_{0}^{a} \sqrt{a^2 - x^2} \; dx$ $\;\;\; \cdots \; (4)$
Let $x = a \sin \theta$ $\;\;\; \cdots \; (5)$
Differentiating equation $(5)$ gives
$dx = a \cos \theta \; d\theta$ $\;\;\; \cdots \; (5a)$
Also, we have from equation $(5)$, $\theta = \sin^{-1} \left(\dfrac{x}{a}\right)$
$\text{When} \begin{cases}
x = 0, & \theta = \sin^{-1} \left(0\right) = 0 \\
x = a, & \theta = \sin^{-1} \left(1\right) = \dfrac{\pi}{4}
\end{cases}$ $\;\;\; \cdots \; (5b)$
$\therefore$ $\;$ In view of equations $(5)$, $(5a)$ and $(5b)$, equation $(4)$ can be written as
$\begin{aligned}
I & = \int \limits_{0}^{\frac{\pi}{2}} \sqrt{a^2 - a^2 \sin^2 \theta} \times a \; \cos \theta \; d \theta \\\\
& = \int \limits_{0}^{\frac{\pi}{2}} a \sqrt{1 - \sin^2 \theta} \times a \; \cos \theta \; d \theta \\\\
& = a^2 \int \limits_{0}^{\frac{\pi}{2}} \cos \theta \times \cos \theta \; d \theta \\\\
& = a^2 \int \limits_{0}^{\frac{\pi}{2}} \cos^2 \theta \; d \theta \\\\
& = a^2 \int \limits_{0}^{\frac{\pi}{2}} \left(\dfrac{1 + \cos 2 \theta}{2}\right) \; d \theta \\\\
& = \dfrac{a^2}{2} \left\{\int \limits_{0}^{\frac{\pi}{2}} d \; \theta + \int \limits_{0}^{\frac{\pi}{2}} \cos 2 \theta \; d \theta \right\} \\\\
& = \dfrac{a^2}{2} \left\{\left[\theta\right]_{0}^{\pi / 2} + \dfrac{1}{2} \left[\sin 2 \theta\right]_{0}^{\pi / 2} \right\} \\\\
& = \dfrac{a^2}{2} \left\{\dfrac{\pi}{2} - 0 + \dfrac{1}{2} \left[\sin \left(\pi\right) - \sin \left(0\right)\right] \right\} \\\\
& = \dfrac{\pi \; a^2}{4} + 0 = \dfrac{\pi \; a^2}{4}
\end{aligned}$
$\therefore$ $\;$ Required area $= 4 \times \dfrac{\pi \; a^2}{4} = \pi a^2$ square units