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.
 A: 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}
A: 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.
A: A slightly different approach: With the change of variables $u=\cos\theta$, you arrive at
$$\int_0^{\pi/2}\sqrt{\sin2\theta}\sin\theta\,d\theta=\sqrt2\int_0^1 u^{1/2}(1-u^2)^{1/4}\,du$$
Another substitution, $\sqrt t=u$, yields
$$\begin{align*}\sqrt2\int_0^1 t^{1/4}(1-t)^{1/4}\left(\frac{1}{2}t^{-1/2}\right)\,dt&=\frac{\sqrt2}{2}\int_0^1 t^{3/4-1}(1-t)^{5/4-1}\,dt\\[1ex]&=\frac{\sqrt2}{2}\mathrm{B}\left(\frac{3}{4},\frac{5}{4}\right)\end{align*}$$
which is equivalent to Leucippus' solution.
