<s>[INST] Prove that $ \int_{0}^{1}\frac{\sqrt{1-x^2}}{1-x^2\sin^2 \alpha}dx = \frac{\pi}{4\cos^2 \frac{\alpha}{2}}$. Prove that $\displaystyle \int_{0}^{1}\frac{\sqrt{1-x^2}}{1-x^2\sin^2 \alpha}dx = \frac{\pi}{4\cos^2 \frac{\alpha}{2}}$.
$\bf{My\; Try}::$ Let $x = \sin \theta$, Then $dx = \cos \theta d\theta$
$\displaystyle = \int _{0}^{1}\frac{\cos \theta }{1-\sin^2 \theta \cdot \sin^2 \alpha}\cdot \cos \theta d\theta = \int_{0}^{1}\frac{\cos ^2 \theta }{1-\sin^2 \theta \cdot \sin ^2 \alpha}d\theta$
$\displaystyle = \int_{0}^{1}\frac{\sec^2 \theta}{\sec^4 \theta -\tan^2 \theta \cdot \sec^2 \theta \cdot \sin^2 \alpha}d\theta = \int_{0}^{1}\frac{\sec^2 \theta }{\left(1+\tan ^2 \theta\right)^2-\tan^2 \theta \cdot \sec^2 \theta\cdot \sin^2 \alpha}d\theta$
Let $\tan \theta = t$ and $\sec^2 \theta d\theta = dt$
$\displaystyle \int_{0}^{1}\frac{1}{(1+t^2)^2-t^2 (1+t^2)\sin^2 \alpha}dt$
Now How can I solve after that
Help Required
Thanks
 [/INST] Substitute $x=\frac{t}{\sqrt{\cos a+t^2}}$. Then
$dx= \frac{\cos a\ dt}{(\cos a+t^2)^{3/2}}$ and
\begin{align}\int_{0}^{1}\frac{\sqrt{1-x^2}}{1-x^2\sin^2 a}dx
=& \int_{0}^{\infty}\frac{\sqrt{\cos a}\ \frac1{t^2} }{(t^2+ \frac1{t^2} )\cos a +(1+\cos^2a)}\overset{t\to 1/t}{dt}\\
 =& \ \frac{\sqrt{\cos a}}2\int_{0}^{\infty}\frac{d(1-\frac1{t} )}{(t-\frac1{t} )^2\cos a +(1+\cos a)^2}\\
 =& \ \frac{\pi}{2(1+\cos a)}\\
\end{align}
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