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.
 A: with $\cos\theta = \tan \frac t2$
\begin{align}\int_{0}^{\pi/2}\frac{\sqrt{\cos \theta}}{1 + \cos^2\theta}d\theta =\frac{1}{2\sqrt2}\int_0^{\pi/2}\sqrt{\tan t}\>dt\overset{x^2=\cot t}=\frac1{\sqrt2}\int_0^\infty \frac{1}{1+x^4}dx
= \frac{\pi}{4}
\end{align}
A: Maybe this identity for ${}_{2}F_{1}$ helps:
$\displaystyle I=\int_{0}^\frac{\pi}{2} \frac{\sqrt{\cos\theta}}{1+\cos^2 \theta}d\theta = \sum_{n=0}^{\infty} (-1)^n\int_{0}^{\frac{\pi}{2}} \cos^{2n+\frac{1}{2}}\theta d\theta = \sum_{n=0}^{\infty} (-1)^n\frac{\sqrt{\pi}\Gamma\left(n+\frac{3}{4}\right)}{2\Gamma\left(n+\frac{5}{4}\right)}$
Now using
$\displaystyle (x)_{n} = \frac{\Gamma(n+x)}{\Gamma(x)}$ and $(1)_{n}=n!$
where $(x)_{n}$ is the rising factorial
$\displaystyle I = \sum_{n=0}^{\infty} (-1)^n\frac{\sqrt{\pi}\Gamma\left(n+\frac{3}{4}\right)}{2\Gamma\left(n+\frac{5}{4}\right)}=  \frac{\sqrt{\pi}\Gamma\left(\frac{3}{4}\right)}{2\Gamma\left(\frac{5}{4}\right)}\sum_{n=0}^{\infty} (-1)^n\frac{\left(\frac{3}{4}\right)_{n}}{\left(\frac{5}{4}\right)_{n}} =\frac{\sqrt{\pi}\Gamma\left(\frac{3}{4}\right)}{2\Gamma\left(\frac{5}{4}\right)}\sum_{n=0}^{\infty} \frac{(1)_{n}\left(\frac{3}{4}\right)_{n}}{\left(\frac{5}{4}\right)_{n}}\frac{(-1)^n}{n!} = \frac{\sqrt{\pi}\Gamma\left(\frac{3}{4}\right)}{2\Gamma\left(\frac{5}{4}\right)}{}_{2}F_{1}\left(1,\frac{3}{4};\frac{5}{4};-1\right)$
With this identity:
$ \displaystyle {}_{2}F_{1}(a,b;a-b+1;-1) = \frac{\Gamma(a-b+1)\Gamma\left(\frac{1}{2}a+1\right)}{\Gamma(a+1)\Gamma\left(\frac{1}{2}a-b+1\right)}$
we have
$\displaystyle I= \frac{\sqrt{\pi}\Gamma\left(\frac{3}{4}\right)\Gamma\left(\frac{5}{4}\right)\Gamma\left(\frac{3}{2}\right)}{2\Gamma\left(\frac{5}{4}\right)\Gamma\left(2\right)\Gamma\left(\frac{3}{4}\right)} = \frac{\sqrt{\pi}\Gamma\left(\frac{3}{2}\right)}{2} =\frac{\pi}{4}$
A: 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,
$$
I = \int_{0}^{\pi/2}\dfrac{\sqrt{\cos \theta}}{1 + \cos^2 \theta}d\theta = \int_{\pi/4}^{0}\dfrac{\sqrt{\tan \alpha}(-\sec^2 \alpha)d\alpha}{(1 + \tan^2 \alpha)\sqrt{1 - \tan^2 \alpha}} = \int_{0}^{\pi/4}\sqrt{\dfrac{\tan \alpha}{1 - \tan^2 \alpha}}d\alpha \quad \Rightarrow 
$$
$$
I = \dfrac{1}{\sqrt{2}}\int_{0}^{\pi/4}\sqrt{\dfrac{\sin(2\alpha)}{\cos(2\alpha)}}d\alpha = \dfrac{1}{2\sqrt{2}}\int_{0}^{\pi/2}\sin^{1/2}\alpha \cos^{-1/2}\alpha d\alpha \quad \Rightarrow
$$
$$
I = \dfrac{1}{4\sqrt{2}}B(1/4,3/4)= \dfrac{1}{4\sqrt{2}}\cdot \dfrac{\pi}{\sin(\pi/4)} = \dfrac{\pi}{4}
$$
A: We have
$$\int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\cos(\theta)}}{1+\cos^{2}(\theta)}d\theta=\int_{0}^{\frac{\pi}{2}}\sum_{n=0}^{\infty}(-1)^{n}\cos^{2n+\frac{1}{2}}(\theta)d\theta$$
$$=\sum_{n=0}^{\infty}(-1)^{n}\int_{0}^{\frac{\pi}{2}}\cos^{2n+\frac{1}{2}}(\theta)d\theta$$
by Fubini/Tonelli theorems. Then using the trigonometric representation of the Beta function
$$\frac{1}{2}\beta\left(\frac{1+n}{2},\frac{1+m}{2}\right)=\int_{0}^{\frac{\pi}{2}}\cos^{n}(x)\sin^{m}(x)dx,\space \text{for}\space n,m>-1$$
we obtain $$\int_{0}^{\frac{\pi}{2}}\cos^{2n+\frac{1}{2}}(\theta)d\theta=\frac{1}{2}\beta\left(n+\frac{3}{4},\frac{1}{2}\right)=\frac{1}{2}\frac{\Gamma\left(n+\frac{3}{4}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma\left(n+\frac{5}{4}\right)}$$
and the integral becomes
$$\int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\cos(\theta)}}{1+\cos^{2}(\theta)}d\theta=\frac{\sqrt{\pi}}{2}\sum_{n=0}^{\infty}(-1)^{n}\frac{\Gamma\left(n+\frac{3}{4}\right)}{\Gamma\left(n+\frac{5}{4}\right)}$$
$$=\frac{\sqrt{\pi}}{2}\left(\frac{5\Gamma\left(\frac{7}{4}\right)\, _2F_1\left(\frac{3}{4},1;\frac{9}{4}; -1\right)\, }{3\, \Gamma \left(\frac{9}{4}\right)}-\frac{4\Gamma\left(\frac{11}{4}\right)\, _2F_1\left(\frac{7}{4},2;\frac{13}{4}; -1\right)\, }{7\, \Gamma \left(\frac{13}{4}\right)}\right)=\frac{\pi}{4}$$
where $_2F_1(a,b;c;x)$ is the hypergeometric function.
