# The Little Monster Integral

**Calculus**Level 5

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 \(\large\alpha+\beta\).