\documentclass[12pt, a4paper]{letter}

\begin{document}

$\int_0^r \sqrt{r^2 - x^2} dx$

let $x = r\sin\theta$

$dx = r\cos\theta d\theta$

$\int_0^\frac{\pi}{2} \sqrt{r^2 - r^2\sin^2\theta} r \cos\theta d\theta$

$ = r^2 \int_0^\frac{\pi}{2} \sqrt{1 - \sin^2\theta} \cos\theta d\theta$

$ = r^2 \int_0^\frac{\pi}{2} \cos^2\theta d\theta$

$\cos2\theta = 2\cos^2\theta - 1$

$\cos^2\theta = \frac{\cos2\theta + 1}{2}$

$ = r^2 \int_0^\frac{\pi}{2} \frac{\cos2\theta + 1}{2} d\theta$

$ = \frac{r^2}{2} \int_0^\frac{\pi}{2} (\cos2\theta + 1) d\theta$

$ = \frac{r^2}{2} (\frac{1}{2}\sin2\theta|_0^\frac{\pi}{2} + \theta|_0^\frac{\pi}{2})$

$ = \frac{r^2}{2} (0 + \frac{\pi}{2})$

$ = \frac{\pi r^2}{4}$

$ \frac{\pi r^2}{4} * 4 = \pi r^2$
\end{document}