TITLE: Finding the definite integral of a trigonometric expression
QUESTION [6 upvotes]: Find the integral of $$ \int_0^{\frac{\pi}{2}}{{\sqrt{\sin(2\theta)}} \cdot \sin(\theta)d\theta}$$
I got $$I=\int_0^\frac{\pi}{4}{\sqrt{\sin(2\theta)} \cdot (\sin(\theta)+\cos(\theta))d\theta}$$
But, I'm stuck here.

REPLY [6 votes]: If you are interested in knowing how to do this without using the Beta function, try the following steps. But I'm not going to write it out in full because it would take too long.
Call the integral $I$. First do integration by parts, and we find that $$I=\int_0^{\frac {\pi}{2}}\frac{\cos 2\theta\cos \theta}{\sqrt{\sin 2\theta}}d\theta$$
Now add this form of $I$ to the original form and get $$2I=\int_0^{\frac {\pi}{2}}\frac{\cos \theta}{\sqrt{\sin 2\theta}}d\theta$$
hence$$4I=\int_0^{\frac {\pi}{2}}\frac{\sqrt{\sin 2\theta}}{\sin\theta}d\theta$$
Now substitute $t=\sqrt{\tan \theta}$ and you end up with a well known integral, featured many times on MSE, requiring a rather tedious partial fraction decomposition, but you get there in the end...
I hope this is sufficient.