The following integral involving inverse trigonometric and trigonometric functions has a gee-whiz closed-form
\[\begin{equation} {\large\int_0^{\Large\frac{\pi}{2}}} {\rm{arccot}}\sqrt{\frac{1+\sin\theta}{\sin\theta}}\,\,d\theta={\large\frac{\pi^{\alpha}}{\beta}} \end{equation}\]
Find the value of \(\large\alpha+\beta\).
by
Anastasiya Romanova
