Area of a Circle $A=\pi r^2$ using integration
To find the area of the circle we simply integrate to find the area between $x=0$ and $x=r$ and multiply by $4$.
Area $$A=4\int^r_0 \sqrt{r^2-x^2}\,dx$$
Using the substitution $x=r \sin\theta$ gives $$dx=r \cos\theta$$ and the integral converts to
$\displaystyle\begin{array}{ll}A&=\displaystyle 4\int^{\pi/2}_0 \sqrt{r^2-r^2\sin^2\theta}.r\cos\theta \,d\theta\\
&=\displaystyle 4\int^{\pi/2}_0 \sqrt{r^2(1-\sin^2\theta)}.r\cos\theta \,d\theta\\
&=\displaystyle4\int^{\pi/2}_0 \sqrt{r^2\cos^2\theta}.r\cos\theta \,d\theta\\
&=\displaystyle 4\int^{\pi/2}_0 r\cos\theta.r\cos\theta \,d\theta\\
&=\displaystyle 4\int^{\pi/2}_0 r^2\cos^2\theta \,d\theta\\
&=\displaystyle 4r^2\int^{\pi/2}_0 {1+\cos 2\theta\over 2}\, d\theta\\
&=\displaystyle 2r^2\left. \left( \theta+{\sin 2\theta\over 2}\right)\right|^{\pi/2}_0\\
&=\displaystyle 2r^2\left( {\pi\over 2} +{\sin\pi\over 2}-0- {\sin 0\over 2}\right)\\
&=\displaystyle\pi r^2
\end{array}$
