Computing $\int_0^{2\pi}\sqrt{1+\sin x}dx$

A formula for $\sin x+\sin y$ may help:

$$1+\sin x = \sin\frac{\pi}{2}+\sin x = 2\left(\sin\left(\frac{\pi}{4}+\frac{x}{2}\right)\right)^2$$

and now it should be easier