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
 A: Here is a much nicer way than my first attempt. Use $x=\sin(\theta)$ and $u=\tan(\theta)$
$$
\begin{align}
&\int_0^1\frac{\sqrt{1-x^2}}{1-x^2\sin^2(\alpha)}\,\mathrm{d}x\\
&=\int_0^{\pi/2}\frac{\cos^2(\theta)}{1-\sin^2(\theta)\sin^2(\alpha)}\,\mathrm{d}\theta\\
&=\int_0^{\pi/2}\frac{\mathrm{d}\theta}{\sec^2(\theta)-\tan^2(\theta)\sin^2(\alpha)}\\
&=\int_0^{\pi/2}\frac{\mathrm{d}\theta}{1+\tan^2(\theta)\cos^2(\alpha)}\\
&=\int_0^{\pi/2}\frac{\mathrm{d}\tan(\theta)}{(1+\tan^2(\theta)\cos^2(\alpha))(1+\tan^2(\theta))}\\
&=\frac1{\sin^2(\alpha)}\int_0^{\pi/2}\left(\frac1{1+\tan^2(\theta)}-\frac{\cos^2(\alpha)}{1+\tan^2(\theta)\cos^2(\alpha)}\right)\,\mathrm{d}\tan(\theta)\\
&=\frac1{\sin^2(\alpha)}\int_0^\infty\left(\frac1{1+u^2}-\frac{\cos^2(\alpha)}{1+u^2\cos^2(\alpha)}\right)\,\mathrm{d}u\\
&=\frac1{\sin^2(\alpha)}\left(\frac\pi2-\frac\pi2\cos(\alpha)\right)\\
&=\frac\pi2\frac1{1+\cos(\alpha)}\\
&=\frac\pi{4\cos^2(\alpha/2)}
\end{align}
$$
A: 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}
A: 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}
A: 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
$$
\begin{aligned}
I&=\int_0^{\infty} \frac{\frac{1}{t^2}}{t^2 \cos ^2 \alpha+\frac{1}{t^2}+\left(1+\cos ^2 \alpha\right)} d t \\
&=\frac{1}{2} \int_0^{\infty} \frac{\left(\cos \alpha+\frac{1}{t^2}\right)-\left(\cos \alpha-\frac{1}{t^2} \right)}{t^2 \cos ^2 \alpha+\frac{1}{t^2}+\left(1+\cos ^2 \alpha\right)} d t \\
&=\frac{1}{2}\left[\int_0^{\infty} \frac{d\left(t \cos \alpha-\frac{1}{t}\right)}{\left(t \cos \alpha-\frac{1}{t}\right)^2+(1+\cos \alpha)^2}+\int_0^{\infty} \frac{d\left(t \cos \alpha+\frac{1}{t}\right)}{\left.\left(t \cos \alpha+\frac{1}{t}\right)^2+(1-\cos \alpha t)\right]}\right] \\
&=\frac{1}{2(1+\cos \alpha)}\left[\tan ^{-1}\left(\frac{t \cos \alpha-\frac{1}{t}}{1+\cos \alpha}\right)\right]_0^{\infty} \\
&=\frac{1}{4 \cos ^2 \frac{\alpha}{2}}\left(\frac{\pi}{2}+\frac{\pi}{2}\right) \\
&=\frac{\pi}{4 \cos ^2 \frac{\alpha}{2}}
\end{aligned}
$$
