# 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 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}