Evaluate $\int x \frac{\sqrt {a^2 - x^2}}{\sqrt{a^2+x^2}} dx$ How to solve $\displaystyle\int x\, \frac{\sqrt {a^2 - x^2}}{\sqrt{a^2+x^2}}\, dx$?
I have tried substituting $x = a \sin(\theta)$ and got
$$a^2\int\frac{\sin(\theta) \cos^2(\theta)}{\sqrt{1+\sin^2(\theta)}}\, d \theta.$$
I'm not sure how to proceed from here.

Note
$$\int\frac{\sin \theta \cos^2 \theta}{\sqrt{1+\sin^2 \theta}} d \theta=-\int\frac{\cos^2 \theta}{\sqrt{2-\cos^2 \theta}} d \cos\theta.$$
Let $u=\cos\theta$ and then
$$\int\frac{\cos^2 \theta}{\sqrt{2-\cos^2 \theta}} d \cos\theta=\int\frac{u^2}{\sqrt{2-u^2}}du.$$
Let $u=\sqrt{2}\sin t$ and then
$$\int\frac{u^2}{\sqrt{2-u^2}}du=\int\frac{2\sin^2t}{\sqrt2\cos t}\sqrt2\cos tdt=\int(1-\cos(2t))dt=\cdots. $$
which you can handle easily.
