# How do I calculate $\int^{\pi/2}_0 \sqrt{\cos(t)\sin(t)} dt$

I need to calculate the following integral: $$\int^{\pi/2}_0 \sqrt{\cos(t)\sin(t)} dt$$ Symbolic computation package gives me an answer: $$\int^{\pi/2}_0 \sqrt{\cos(t)\sin(t)} dt = \frac{\Gamma(\frac{3}{4})^2}{\sqrt{\pi}}$$ I don't really see any property of Gamma function which I could use in here. I thought of using exp function, but it doesn't really let me move towards Gamma.

$$\int^{\pi/2}_0 \sqrt{\cos(t)\sin(t)} dt = \int^{\pi/2}_0 \cos(t)e^{\ln(\frac{1}{2}\sin(2t))} dt$$

What other approach can I use to solve this integral?

This problem belongs to my homework assignment.

$$\int_{0}^{\pi/2}\sqrt{\sin x\cos x}\,dx \stackrel{x=\arcsin t}{=} \int_{0}^{1}t^{1/2}(1-t^2)^{-1/4}\,dt\stackrel{t=\sqrt{u}}{=}\frac{1}{2}\int_{0}^{1}u^{-1/4}(1-u)^{-1/4}\,du$$ and by Euler's Beta function the RHS equals $\frac{\Gamma\left(\frac{3}{4}\right)^2}{2\Gamma\left(\frac{3}{2}\right)}=\frac{1}{\sqrt{\pi}}\Gamma\left(\frac{3}{4}\right)^2$, which is clearly related to the lemniscate constant. An equivalent form of the LHS is $$\frac{1}{\sqrt{2}}\int_{0}^{\pi/2}\sqrt{\sin(2x)}\,dx=\sqrt{2}\int_{0}^{\pi/4}\sqrt{\sin(2x)}\,dx = \frac{1}{\sqrt{2}}\int_{0}^{\pi/2}\sqrt{\sin x}\,dx$$ or $$\frac{1}{\sqrt{2}}\int_{0}^{\pi/2}\sqrt{\cos x}\,dx = \frac{1}{\sqrt{2}}\int_{0}^{+\infty}\frac{du}{(1+u^2)^{5/4}}$$ and an efficient numerical computation is allowed by the relations between $\Gamma\left(\frac{1}{4}\right)$, complete elliptic integrals of the first kind and the AGM mean: $$\boxed{\int_{0}^{\pi/2}\sqrt{\sin x\cos x}\,dx=\frac{1}{\sqrt{\pi}}\,\Gamma\left(\frac{3}{4}\right)^2=\frac{2\pi^{3/2}}{\Gamma\left(\frac{1}{4}\right)^2}=\text{AGM}\left(1,\frac{1}{\sqrt{2}}\right).}$$

You can use the Beta function identity: $$B(x, y) := \int_{0}^{1}t^{x-1}(1-t)^{y-1}= \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)} = 2\int_{0}^{\pi/2}\sin^{2x-1}\theta\cos^{2x-1}\theta d\theta.$$

Hint: You can reach to gamma function by these steps:

$:\displaystyle \int_{0}^{\pi/2}\sqrt{cos(t) sin(t)}dt= \int_{0}^{\pi/2}\sqrt{\dfrac{sin(2t)}{2} }dt=\int_{0}^{\pi}\sqrt{\dfrac{sin(t)}{2} }\cdot dt/2$

$:\displaystyle \int_0^x \sqrt{\sin t}\ \text dt=\sqrt{\frac{2}{\pi }} \Gamma\left(\frac{3}{4}\right)^2-2 \text{EllipticE}\left[\frac{1}{4} (\pi -2 x),2\right],\quad 0\leq x \leq \pi,$

Where EllipticE$[ϕ,m]$ is a special function studied by Legendre and you will find many of its properties here. Substitute $x$ by $\pi$.

P.S. I used the second equation from Olivier's answer because it is consice.

Applying product to sum formulas: $$\int \sqrt{\cos (t)\sin (t)}dt=\int\dfrac{\sqrt{\sin (2t)}}{\sqrt 2}dt=\dfrac{1}{\sqrt 2}\int\sqrt{\sin(2t)}dt$$

Let $u=2t\rightarrow dt=\dfrac 12du$.

$$\dfrac 12\int\sqrt{\sin(u)}du=\dfrac 12\int\sqrt{2\cos\left(\dfrac{2u-\pi}4\right)-1}du=\dfrac 12\int\sqrt{1-2\sin^2\left(\dfrac{2u-\pi}4\right)}$$

Let $v=\dfrac{2u-\pi}4\rightarrow du=2dv$

$$\dfrac 12\int\sqrt{1-2\sin^2\left(\dfrac{2u-\pi}4\right)}=\int\sqrt{1-2\sin^2(v)}dv=E(v|2)$$

Where $E$ is the incomplete Elliptic Integral of the Second Kind.

Now substitute back in $u$, and substitute back in $t$ for the answer of:

$$\int\sqrt{\cos(x)\sin(x)}=\dfrac{E\left(\dfrac{4x-\pi}{4}\Big{|}2\right)}{\sqrt2}+C$$

You can rationalize that if that doesn't seem proper.