# 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.

• You could also first let $t=x^2$ and then $s=\sqrt{(a^2-t)/(a^2+t)}$. – Hans Lundmark Feb 22 '18 at 17:57

Set $x^2=a^2\cos2t,0\le2t\le\dfrac\pi2$ and $\cos2t=\cdots\ge0$
$\implies x\ dx=-\sin2t\ dt$ and $\sin2t=+\sqrt{1-\cos^22t}=?$
$$\sqrt{\dfrac{a^2-x^2}{a^2+x^2}}=+\tan t$$ and $0\le t\le\dfrac\pi4,\tan t\ge0$
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.