Evaluate the following integrals: ^ a _0x a^2-x^2 a^2+x^2 dx

Question

Evaluate the following integrals:
$\int\limits^{\text{a}}_0\text{x}\sqrt{\frac{\text{a}^2-\text{x}^2}{\text{a}^2+\text{x}^2}}\text{ dx}$

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Answer

Let $\text{I}=\int^\limits{\text{a}}_0\text{x}\sqrt{\frac{\text{a}^2-\text{x}^2}{\text{a}^2+\text{x}^2}}\text{ dx}$
Consider, $\text{x}^2=\text{a}^2\cos2\theta$
$\Rightarrow2\text{xdx}=-2\text{a}^2\sin2\theta\text{ d}\theta$
$\Rightarrow\text{xdx}=-\text{a}^2\sin2\theta\text{ d}\theta$
When, $\text{x}\rightarrow0;\ \theta\rightarrow\frac{\pi}{4}$ and $\text{x}\rightarrow\text{a};\ \theta\rightarrow0$
Now, integral becomes,
$\text{I}=\int^\limits0_\frac{\pi}{4}-\text{a}^2\sin2\theta\sqrt{\frac{\text{a}^2-\text{a}^2\cos2\theta}{\text{a}^2+\text{a}^2\cos2\theta}}\text{ d}\theta$
$=\int^\limits0_\frac{\pi}{4}-\text{a}^2\sin2\theta\tan\theta\text{ d}\theta$
$=\text{a}^2\int^\limits{\frac{\pi}{4}}_02\sin\theta\cos\theta\frac{\sin\theta}{\cos\theta}\text{ d}\theta$
$=\text{a}^2\int^\limits{\frac{\pi}{4}}_02\sin\theta\text{ d}\theta$
$=\text{a}^2\int^\limits{\frac{\pi}{4}}_0\big[1-\cos\theta\big]\text{d}\theta$
$=\text{a}^2\Big[\theta-\frac{\sin2\theta}{2}\Big]^{\frac{\pi}{4}}_0$
$=\text{a}^2\Big[\frac{\pi}{4}-\frac{1}{2}\Big]$

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