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

Here is another method to solve the problem by using residue. Suppose $\alpha\in(0,\pi/2)$. Using $x=\sin(\theta)$ and $u=\tan(\theta), z=e^{i\theta}$, one has
\begin{eqnarray}
&&\int_0^1\frac{\sqrt{1-x^2}}{1-x^2\sin^2(\alpha)}\,\mathrm{d}x\\
&=&\int_0^{\pi/2}\frac{\cos^2(\theta)}{1-\sin^2(\theta)\sin^2(\alpha)}\,\mathrm{d}\theta\\
&=&\int_0^{\pi/2}\frac{\frac{1+\cos(2\theta)}{2}}{1-\frac{1-\cos(2\theta)}{2}\sin^2(\alpha)}\,\mathrm{d}\theta\\
&=&\frac14\int_0^{2\pi}\frac{1+\cos(\theta)}{2-(1-\cos(\theta))\sin^2(\alpha)}\,\mathrm{d}\theta\\
&=&\frac14\int_{|z|=1}\frac{1+\frac{z+z^{-1}}{2}}{2-(1-\frac{z+z^{-1}}{2})\sin^2(\alpha)}\frac{1}{iz}\,\mathrm{d}z\\
&=&\frac1{4i}\int_{|z|=1}\frac{(z+1)^2}{z[4+(z-1)^2\sin^2(\alpha)]}\,\mathrm{d}z\\
&=&\frac1{4i\sin^2(\alpha)}\int_{|z|=1}\frac{(z+1)^2}{z(z+\tan^2(\alpha/2))(z+\cot^2(\alpha/2))}\,\mathrm{d}z\\
&=&\frac1{4i\sin^2(\alpha)}2\pi i\bigg[\text{Res}(\frac{(z+1)^2}{(z+\tan^2(\alpha/2))(z+\cot^2(\alpha/2))},z=0)+\text{Res}(\frac{(z+1)^2}{z(z+\cot^2(\alpha/2))},z=-\tan^2(\alpha/2)\bigg]\\
&=&\frac{\pi}{4\cos^2(\alpha/2)}.
\end{eqnarray}
