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

Letting $x=\sin \theta$ yields
$$
\begin{aligned}
I &=\int_0^{\frac{\pi}{2}} \frac{d t}{\left(1+t^2\right)^2-\left(1+t^2\right) t^2 \sin ^2 \alpha} \\
&=\int_0^{\infty} \frac{d x}{t^4 \cos ^2 \alpha+\left(1+\cos ^2 \alpha\right) t^2+1}
\end{aligned}
$$
Dividing both denominator and numerator of the integrand gives
$$
\begin{aligned}
I&=\int_0^{\infty} \frac{\frac{1}{t^2}}{t^2 \cos ^2 \alpha+\frac{1}{t^2}+\left(1+\cos ^2 \alpha\right)} d t \\
&=\frac{1}{2} \int_0^{\infty} \frac{\left(\cos \alpha+\frac{1}{t^2}\right)-\left(\cos \alpha-\frac{1}{t^2} \right)}{t^2 \cos ^2 \alpha+\frac{1}{t^2}+\left(1+\cos ^2 \alpha\right)} d t \\
&=\frac{1}{2}\left[\int_0^{\infty} \frac{d\left(t \cos \alpha-\frac{1}{t}\right)}{\left(t \cos \alpha-\frac{1}{t}\right)^2+(1+\cos \alpha)^2}+\int_0^{\infty} \frac{d\left(t \cos \alpha+\frac{1}{t}\right)}{\left.\left(t \cos \alpha+\frac{1}{t}\right)^2+(1-\cos \alpha t)\right]}\right] \\
&=\frac{1}{2(1+\cos \alpha)}\left[\tan ^{-1}\left(\frac{t \cos \alpha-\frac{1}{t}}{1+\cos \alpha}\right)\right]_0^{\infty} \\
&=\frac{1}{4 \cos ^2 \frac{\alpha}{2}}\left(\frac{\pi}{2}+\frac{\pi}{2}\right) \\
&=\frac{\pi}{4 \cos ^2 \frac{\alpha}{2}}
\end{aligned}
$$
