# Problem 20 **

Use the substitution x = r \sin \theta. Then \mathrm dx = r \cos \theta \mathrm d\theta so that:

\begin{align*}\int \sqrt {r^2 - x^2}\mathrm dx & = r \int \cos \theta \sqrt {r^2 - r^2 \sin^2 \theta} \mathrm d\theta \\ & = r^2 \int \cos^2 \theta \mathrm d\theta \\ & = \frac {r^2} 2 \int (1 + \cos 2\theta) \mathrm d\theta \\ & = \frac {r^2} 2 \left(\theta + \frac 1 2 \sin 2 \theta\right) + C \\ & = \frac {r^2} 2 \left(\arcsin \frac x r + \sin \arcsin \frac x r \cos \arcsin \frac x r\right) + C \\ & = \frac {r^2} 2 \left(\arcsin \frac x r + \frac x r \sqrt {1 - \frac {x^2} {r^2}}\right) + C \\ & = \frac 1 2 \left(r^2 \arcsin \frac x r + x \sqrt {r^2 - x^2}\right) + C \end{align*}

The only hard bit now is the first term in the brackets. Say:

\displaystyle x = \sin \theta

with \displaystyle -\frac \pi 2 < \theta < \frac \pi 2

then:

\displaystyle \frac x {\sqrt {1 - x^2}} = \frac {\sin \theta} {\cos \theta} = \tan \theta

So, noting again that \displaystyle -\frac \pi 2 < \theta < \frac \pi 2:

\displaystyle \theta = \arctan \left(\frac x {\sqrt {1 - x^2}}\right)

giving:

\displaystyle \arcsin x = \arctan \left(\frac x {\sqrt {1 - x^2}}\right)

So:

\displaystyle \arcsin \frac x r = \arctan \left(\frac x {r \sqrt {1 - \frac {x^2} {r^2}}}\right) = \arctan \left(\frac x {\sqrt {r^2 - x^2}}\right)

Giving:

\displaystyle \int \sqrt {r^2 - x^2} \mathrm dx = \frac 1 2 \left(r^2 \arctan \left(\frac x {\sqrt {r^2 - x^2}}\right) + x \sqrt {r^2 - x^2}\right) + C

Note that the curve y = \sqrt {r^2 - x^2} for 0 \le x \le r traces out the quarter circle centred at the origin with radius r so:

\begin{align*} A &= 4 \int_0^r \sqrt{r^2 - x^2} \mathrm dx \\ & = 2 \left[r^2 \arcsin \frac x r + x \sqrt {r^2 - x^2}\right]_0^r \\ & = 2 \left(r^2 \arcsin 1 + r \sqrt {r^2 - r^2} - r^2 \arcsin 0 - 0 \sqrt {r^2 - 0^2}\right) \\\ & = 2 r^2 \arcsin 1 \\ & = \pi r^2\end{align*}