I was wondering about the integral $\int_{0}^{a}{\sqrt{\sin{x}}}dx$ and after some time I arrived at an indefinite integral of $-2E(\frac{\pi}{4}-\frac{x}{2}|2)+c$ where E is the elliptic integral of second kind.
By the fundamental theorem of Calculus, I would need to evaluate the integral when $x=0$, in other words $E(\frac{\pi}{4}|2)$, which is
$$ \int_0^{\pi/4}\sqrt{1-2\sin^{2}{x}}dx=2\sqrt{\frac{2}{\pi}}\Gamma\left(\frac{3}{4}\right)^2 $$
by wolfram alpha.
I tried many methods like king’s property but all methods fail to evaluate this.
Prove a special case of Complete Elliptic Integral of the Second Kind This link is slightly related to my question, but its about the complete elliptic integral. I still don’t get how to evaluate the integral with bounds $0$ and $\frac{\pi}{4}$.
Could anyone please show me how this integral is evaluated to give this value? Thank you so much!
Edit: I realised that the integral could simplify to the square root of cosine. Then the integral from $\frac{\pi}{4}$ to $\frac{\pi}{2}$ would be $i$ times the integral from $0$ to $\frac{\pi}{4}$, so the value is easily derived from the one in the link provided.
But could someone explain in detail why the integral gives this value? Thank you!