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!


1 Answer 1


$$\int\sqrt{1-2\sin^{2}{x}}\,dx=E(x|2)$$ $$\int_0^{\frac \pi 4}\sqrt{1-2\sin^{2}{x}}\,dx=\frac{1}{\sqrt{2}} \left(2 E\left(\frac{1}{2}\right)-K\left(\frac{1}{2}\right)\right)$$ $$E\left(\frac{1}{2}\right)=\frac{4 \pi ^{3/2}}{\Gamma \left(-\frac{1}{4}\right)^2}+\frac{\Gamma \left(\frac{3}{4}\right)^2}{2 \sqrt{\pi }}$$ $$K\left(\frac{1}{2}\right)=\frac{8 \pi ^{3/2}}{\Gamma \left(-\frac{1}{4}\right)^2}$$

Just continue the simplifications


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