How do you integrate the reciprocal of square root of cosine? I encountered this integral in physics and got stuck. 
$$\int_{0}^{\Large\frac{\pi}{2}} \dfrac{d\theta}{\sqrt{\cos \theta}}.$$  

Substituting $y=\cos{\theta}$
\begin{align*}
  \int_{0}^{\frac{\pi}{2}} \dfrac{d\theta}{\sqrt{\cos \theta}} &= \int_{0}^{1} \, \frac{1}{\sqrt{y\, \left(1-y^2\right)}}\, dy \\
  &= \frac{1}{2}\int_{0}^{1} \, t^{-3/4}\left(1-t\right)^{-1/2}\, dt \tag{where $t=y^2$} \\
  &= \frac{1}{2}\mathrm{B}\left(\frac{1}{4}, \frac{1}{2}\right) \\
  &= \frac{\sqrt\pi}{2}\frac{\Gamma\left(1/4\right)}{\Gamma\left(3/4\right)} \approx 2.62205755429212
\end{align*}
Look up Beta and Gamma functions.
