<s>[INST] How to prove $\int_{0}^{\pi/2}\frac{\sqrt{\cos \theta}}{1 + \cos^2\theta}d\theta = \frac{\pi}{4}$ Prove that
$$
\int_{0}^{\pi/2}\frac{\sqrt{\cos \theta}}{1 + \cos^2\theta}d\theta = \frac{\pi}{4}
$$
My attempt :I tried to use the beta function, but I couldn't.
 [/INST] In search of a simple solution and after many attempts, I found this solution. Let $\cos \theta = \tan \alpha$ and $d\theta = -\dfrac{\sec^2 \alpha \ d\alpha}{\sqrt{1 - \tan \alpha}}$. Thus,
$$
I = \int_{0}^{\pi/2}\dfrac{\sqrt{\cos \theta}}{1 + \cos^2 \theta}d\theta = \int_{\pi/4}^{0}\dfrac{\sqrt{\tan \alpha}(-\sec^2 \alpha)d\alpha}{(1 + \tan^2 \alpha)\sqrt{1 - \tan^2 \alpha}} = \int_{0}^{\pi/4}\sqrt{\dfrac{\tan \alpha}{1 - \tan^2 \alpha}}d\alpha \quad \Rightarrow 
$$
$$
I = \dfrac{1}{\sqrt{2}}\int_{0}^{\pi/4}\sqrt{\dfrac{\sin(2\alpha)}{\cos(2\alpha)}}d\alpha = \dfrac{1}{2\sqrt{2}}\int_{0}^{\pi/2}\sin^{1/2}\alpha \cos^{-1/2}\alpha d\alpha \quad \Rightarrow
$$
$$
I = \dfrac{1}{4\sqrt{2}}B(1/4,3/4)= \dfrac{1}{4\sqrt{2}}\cdot \dfrac{\pi}{\sin(\pi/4)} = \dfrac{\pi}{4}
$$
</s>