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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. $$ \int x \sqrt{\frac {a^2 - x^2} { a^2 + x^2 }} dx $$

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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.

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Thanks! That's better :D – Sri Krishna Apr 24 '13 at 7:47
@SriKrishna You might be slightly bothered by the different signs for the integral, but the limits will take care of the signs. So be careful with the limits. – user17762 Apr 24 '13 at 7:49

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 $$ \begin{eqnarray*} \int \frac{\sqrt{t}}{\left( 1+t\right) ^{2}}dt &=&2\int \frac{u^{2}}{\left( 1+u^{2}\right) ^{2}}\,du \\ &=&2\int -\frac{1}{\left( 1+u^{2}\right) ^{2}}+\frac{1}{1+u^{2}}\,du.\tag{3} \end{eqnarray*} $$

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