Need help solving $\int x \sqrt{\frac {a^2 - x^2} { a^2 + x^2 }} dx$ I have a complicated integral to solve. Can someone provide a better way to solve it than what i did - dividing by a inside the root, and then putting $ t = x / a $, and then putting $t^2 = \cos \theta $ and so many other substitutions.
$$
  \int x \sqrt{\frac {a^2 - x^2} { a^2 + x^2 }} dx
$$
 A: Let $x^2=a^2\cos(y)$. We then have $2xdx = -a^2 \sin(y) dy$. The integral then becomes
\begin{align}
I & = \int x \sqrt{\dfrac{a^2-x^2}{a^2+x^2}} dx = -\int \dfrac{a^2}2 \sin(y) \sqrt{\dfrac{1-\cos(y)}{1+\cos(y)}}dy = -\dfrac{a^2}2 \int \sin(y) \tan(y/2) dy\\
& = -a^2\int \sin^2(y/2)dy
\end{align}
I trust you can finish it from here.
A: We can use the algebraic substitution:
$$t=\frac{a^{2}-x^{2}}{a^{2}+x^{2}}.\tag{0} $$
We have that
$$\begin{eqnarray*}
I &=&\int x\sqrt{\frac{a^{2}-x^{2}}{a^{2}+x^{2}}}dx=-\int \sqrt{t}\frac{a^{2}
}{\left( 1+t\right) ^{2}}dt\tag{1} \\
&=&a^{2}\frac{\sqrt{t}}{1+t}-a^{2}\arctan \sqrt{t}+C \\
&=&\frac{a^{2}+x^{2}}{2}\sqrt{\frac{a^{2}-x^{2}}{a^{2}+x^{2}}}-a^{2}\arctan 
\sqrt{\frac{a^{2}-x^{2}}{a^{2}+x^{2}}}+C,
\end{eqnarray*}$$
because the integral in $t$ $(1)$ can be evaluated by using another algebraic substitution, $$u^{2}=t,\tag{2}$$ and expanding into partial fractions the resulting integrand
$$
\begin{eqnarray*}
\int \frac{\sqrt{t}}{\left( 1+t\right) ^{2}}dt &=&2\int \frac{u^{2}}{\left(
1+u^{2}\right) ^{2}}\,du \\
&=&2\int -\frac{1}{\left( 1+u^{2}\right) ^{2}}+\frac{1}{1+u^{2}}\,du.\tag{3}
\end{eqnarray*}
$$
