Integral of $\sqrt{R^2-x^2}$ Is there any way to compute the integral
\begin{equation}
\int_{-a}^{a} \sqrt{R^2-x^2} \,\mathrm{d}x, \qquad 0<a<R
\end{equation}
without using trig substitution or integration by parts?
I'm thinking to relate this to area of circle, but I couldn't find the relationship.
Thank you.
 A: Without trigonometric substitution, integration by parts, or appeal to the geometric interpretation of the integral, we need to devise some quite fortuitous manipulations.  We proceed, therefore, and write 
$$\begin{align}
\sqrt{R^2-x^2}&=\frac12 \left(\sqrt{R^2-x^2}+\sqrt{R^2-x^2}\right)\\\\
&=\frac12 \left(\sqrt{R^2-x^2}+\sqrt{R^2-x^2}\right)+\frac{x^2}{\sqrt{R^2-x^2}}-\frac{x^2}{\sqrt{R^2-x^2}}\\\\
=&\frac12 \left(\sqrt{R^2-x^2}-\frac{x^2}{\sqrt{R^2-x^2}}\right)+\frac12\left(\sqrt{R^2-x^2}+\frac{x^2}{\sqrt{R^2-x^2}}\right)\\\\
&=\frac12 \frac{d\left(x\sqrt{R^2-x^2}\right)}{dx}+\frac12\left(R^2-x^2\right)\left(\frac{1}{\sqrt{R^2-x^2}}+\frac{x^2}{(R^2-x^2)^{3/2}}\right)\\\\
&=\frac12 \frac{d\left(x\sqrt{R^2-x^2}\right)}{dx}+\frac12\left(R^2-x^2\right)\frac{d\left(x/\sqrt{R^2-x^2}\right)}{dx}\\\\
&=\frac12 \frac{d\left(x\sqrt{R^2-x^2}\right)}{dx}+\frac12R^2\left(\frac{1}{1+\left(\frac{x}{\sqrt{R^2-x^2}}\right)^2}\right)\frac{d\left(x/\sqrt{R^2-x^2}\right)}{dx}\\\\
&=\frac{d}{dx}\left(\frac12 x\sqrt{R^2-x^2}+\frac12 R^2 \arctan\left(\frac{x}{\sqrt{R^2-x^2}}\right) \right)
\end{align}$$
Therefore, we arrive at 
$$\begin{align}
\int_{-a}^a\sqrt{R^2-x^2}\,dx&=2\int_0^a \sqrt{R^2-x^2}\,dx\\\\
&=\left.\left( x\sqrt{R^2-x^2}+ R^2 \arctan\left(\frac{x}{\sqrt{R^2-x^2}}\right) \right)\right|_0^a\\\\
&= a\sqrt{R^2-a^2}+ R^2 \arctan\left(\frac{a}{\sqrt{R^2-a^2}}\right)
\end{align}$$
A: If you want to use trigonometric identity at the last moment you will find $\sec^3\theta$ for what you have to integration by parts. Let's start :
$$\int_{-a}^{a}\sqrt{R^2-x^2} \ \mathrm dx$$
Let $x=R\tan\theta$ so $\mathrm dx=R\sec^2\theta \ \mathrm d\theta$
$$=\int_{x=-a}^{x=a}R\sqrt{1-\tan^2\theta}R\sec^2\theta \ \mathrm d\theta$$
$$=R^2\int_{x=-a}^{x=a}\sec^3\theta \ \mathrm d\theta$$
$$=R^2\int_{x=-a}^{x=a}\sec\theta\cdot\sec^2\theta \ \mathrm d\theta$$
If you do integration by parts then you will notice you will get recursive integration, I am just taking first part, here you get that $\sec\theta\tan\theta-\int\sec^3\theta\mathrm d\theta+\int \sec\theta\mathrm d\theta$ (ignoring $\sec^3\theta$ for simplicity) and integrand of $\sec\theta$ with respect to theta is $\ln|\sec\theta+\tan\theta|+C$
$$=R^2(\sec\theta\tan\theta|_{\theta=\tan^{-1}(-\dfrac{a}{R})}^{{\theta=\tan^{-1}(\dfrac{a}{R})}}+\ln|\sec\theta+\tan\theta|_{\theta=\tan^{-1}(-\dfrac{a}{R})}^{{\theta=\tan^{-1}(\dfrac{a}{R})}})$$
A: I can't help with the integration part but can sure help with your thought about relating this to a circle.
Think about a circle inside a sphere, if the radius of the sphere is R, the distance of the circle from the center of the sphere is x then;
Radius of the circle is $\sqrt{R^2 - x^2}$
You can do the integration that way too but, this doesn't help much.
