To prove that $\int_0^{\pi/2}\sqrt{\cos(\theta)}d\theta=\frac{(2.\pi)^{3/2}}{16.[\Gamma(5/4)]^2}$, I've already done:
$\int_0^{\pi/2}\sqrt{\cos(\theta)}d\theta=\int_0^{\pi/2}{\cos(\theta)}^{2.(3/4)-1}.{\sin (\theta)}^{2.(1/2)-1}d\theta=\frac{1}{2}.B(3/4,1/2)=\frac{1}{2}.\Gamma(3/4).\Gamma(1/2).[\Gamma(5/4)]^{-1}$
but I don't know how to continue from this to reach $\frac{(2.\pi)^{3/2}}{16.[\Gamma(5/4)]^2}$.
By the first comment, I get it now. Here is the solution:
$\int_0^{\pi/2}\sqrt{\cos(\theta)}d\theta=\int_0^{\pi/2}{\cos(\theta)}^{2.(3/4)-1}.{\sin (\theta)}^{2.(1/2)-1}d\theta=\frac{1}{2}.B(3/4,1/2)=\frac{1}{2}.\Gamma(3/4).\Gamma(1/2).[\Gamma(5/4)]^{-1} = \frac{1}{2}.[\Gamma(3/4).\Gamma(1/4)].[\Gamma(1/4)]^{-1}.\sqrt{\pi}.[\Gamma(5/4)]^{-1}=\frac{1}{2}.\frac{\pi}{\sin(\pi/4)}.[4.\Gamma(5/4)]^{-1}.\sqrt{\pi}.[\Gamma(5/4)]^{-1}= \frac{1}{8}.\sqrt{2}.\pi^{3/2}.[\Gamma(5/4)]^{-2}=\frac{1}{16}.2^{3/2}.\pi^{3/2}.[\Gamma(5/4)]^{-2}$.
Best Answer
I'll spell out what @Mindlack meant. In general $\int_0^{\pi/2}\cos^{2a-1}\theta d\theta=\tfrac12\operatorname{B}(a,\,\tfrac12)=\frac{\sqrt{\pi}\Gamma(a)}{2\Gamma(a+\tfrac12)}$ in terms of Beta and Gamma functions. The case $a=\tfrac34$ obtains your integral as $\frac{\sqrt{\pi}\Gamma(\tfrac34)}{2\Gamma(\tfrac54)}$. The reflection formula $\Gamma(s)\Gamma(1-s)=\pi\csc\pi s$ gives $\Gamma(\tfrac14)\Gamma(\tfrac34)=\pi\sqrt{2}$ at $s=\tfrac14$, so the integral is$$\frac{\sqrt{2\pi^3}}{2\Gamma(\tfrac14)\Gamma(\tfrac54)}=\frac{\sqrt{2\pi^3}}{8\Gamma(\tfrac54)^2}=\frac{(2\pi)^{3/2}}{16\Gamma(\tfrac54)^2},$$since $\Gamma(s)=\frac1s\Gamma(s+1)$, again at $s=\tfrac14$.