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.

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 $\ln|\sec\theta+\tan\theta|+C$
$$=R^2(\sec\theta\tan\theta|_{\theta=\tan^{-1}(-\dfrac{a}{R})}^{{\theta=\tan^{-1}(\dfrac{a}{R})}}+\ln|\sec\theta+\tan\theta|_{\theta=\tan^{-1}(-\dfrac{a}{R})}^{{\theta=\tan^{-1}(\dfrac{a}{R})}})$$
