Finding the definite integral of a trigonometric expression 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.

The integral
$$I = \int_{0}^{\pi/2} \sqrt{\sin(2\theta)} \, \sin\theta \, d\theta$$
is evaluated by making use of the Beta function. This is seen as follows.
\begin{align}
I &= \int_{0}^{\pi/2} \sqrt{\sin(2\theta)} \, \sin\theta \, d\theta \\
&= \sqrt{2} \, \int_{0}^{\pi/2} \sin^{3/2}(\theta) \, \cos^{1/2}(\theta) \, d\theta \\
&= \sqrt{2} \cdot \frac{1}{2} \, B\left(\frac{3}{4}, \frac{1}{4}\right) \\
&= \frac{\Gamma\left(\frac{1}{4}\right) \, \Gamma\left(\frac{3}{4}\right)}{4 \, \sqrt{2}} = \frac{\pi}{4}.
\end{align}
