How to evaluate the integral: $\int_0^\frac{\pi}{2} \cos(x)\sqrt{\cos(x)} \,dx$

Question

$$\int_0^\frac{\pi}{2} \cos(x)\sqrt{\cos(x)} \,dx$$ I've been trying to find a way to integrate this function for a while.
From my research I think this should reduce to an elliptic integral but I can't seem to find a way to reduce it to one of the three canonical forms.

I found this article which might be useful. I've tried to follow some of the steps but I think it's out of my sphere of knowledge at the moment.
So contributions, even if very little, are much appreciated!

Answer 1

I think the easiest way to solve it is by noticing that we can rewrite $\sqrt{\cos x} $ as $ \cos^{\frac12}x$. Thus we have to find $$I=\int_0^\frac{\pi}{2} \cos ^{\frac32}x \mathrm dx $$ Now what about using Beta function? For $m, n > -1$:$$\int_0^\frac{\pi}{2} \sin ^mx \cos ^nx \mathrm dx =\frac12 B \left(\frac{m+1}{2}, \frac{n+1}{2} \right) $$ Now just set $m=0$ and $n=\frac32$. $$I=\frac12B\left(\frac12, \frac54\right)$$ Now using some useful identities such as the relation with the gamma function $$B(x, y) =\frac{\Gamma(x) \Gamma(y)} {\Gamma(x+y)} $$ and that $\Gamma(1+x)=x\Gamma(x)$, $\Gamma\left(\frac12\right)=\sqrt{\pi} $ gives that $$I=\frac12 \frac{\Gamma\left(\frac12\right)\Gamma\left(\frac54 \right)} {\Gamma\left(\frac74 \right)} =\frac12\frac{\sqrt{\pi} \frac14\Gamma\left(\frac1 4 \right)}{\frac34\Gamma\left(\frac34\right)}$$ Additionally one can use gamma reflection formula here if it's desired.

Answer 2

Indeed it is an elliptic integral: $$ I=\int_{0}^{\pi/2}\left(\cos x\right)^{3/2}\,dx\stackrel{x\to \pi/2-x}{=}\int_{0}^{\pi/2}\left(\sin x\right)^{3/2}\,dx\stackrel{x\mapsto \arcsin u}{=}\int_{0}^{1}u^{3/2}(1-u^2)^{-1/2}\,du$$ can be written as $2\int_{0}^{1}\frac{u^4}{\sqrt{1-u^4}}\,du $ or as $$ \frac{1}{2}\int_{0}^{1}v^{1/4}(1-v)^{-1/2}\,dv = \tfrac{1}{2}\,B(5/4,1/2)=\frac{\Gamma\left(\tfrac{5}{4}\right)\Gamma\left(\tfrac{1}{2}\right)}{2\,\Gamma\left(\tfrac{7}{4}\right)}=\frac{\Gamma\left(\tfrac{1}{4}\right)^2}{6\sqrt{2\pi}} $$ due to the properties of the Beta and Gamma functions. On the other hand it also equals $$\frac{\sqrt{2}}{3}\int_{0}^{\pi/2}\frac{d\theta}{\sqrt{1-\tfrac{1}{2}\cos^2\theta}} = \frac{\sqrt{2}}{3}\,K\left(\tfrac{1}{2}\right)=\frac{\pi}{3\,\text{AGM}\left(1,\sqrt{2}\right)}$$ according to Mathematica's notation for the elliptic modulus and the relation between $K$ and the arithmetic-geometric mean. This tour gives a nice relation between a trigonometric integral, a special value for the $\Gamma$ function, a complete elliptic integral of the first kind, the $\text{AGM}$ mean and two hypergeometric series:

$$ 6I=\frac{\Gamma\left(\tfrac{1}{4}\right)^2}{\sqrt{2\pi}}=2\sqrt{2}\,K\left(\tfrac{1}{2}\right)=\frac{2\pi}{\text{AGM}(1,\sqrt{2})} = \frac{1}{\pi\sqrt{2}}\sum_{n\geq 0}\frac{\binom{2n}{n}^2}{32^n}=\sum_{n\geq 0}\frac{6\binom{2n}{n}}{(4n+5)4^n}.$$ This incredible chain of identities led me to ponder the inclusion of the lemniscate constant among the "classical" mathematical constants.

Answer 3

Others have shown you certain evaluations which involve the $\beta$ function, but I'll show you how this general formula is derived. $$I(a,b)=\int_0^{\pi/2}\sin^a(t)\cos^b(t)dt$$ $$u=\sin^2(t)\implies du=2\sin(t)\cos(t)dt\\\implies dt=\frac1{2}u^{-1/2}(1-u)^{-1/2}du$$ I'm sure you can work out the change of the bounds by yourself $$\therefore I(a,b)=\frac1{2}\int_0^1u^{a/2}(1-u)^{b/2}u^{-1/2}(1-u)^{-1/2}du$$ $$\therefore I(a,b)=\frac1{2}\int_0^1u^{(a-1)/2}(1-u)^{(b-1)/2}du$$ $$\therefore I(a,b)=\frac1{2}\int_0^1u^{(a+1)/2-1}(1-u)^{(b+1)/2-1}du$$ And now we recall that $$\beta(a,b)=\int_0^1 t^{a-1}(1-t)^{b-1}dt$$ Which gives $$I(a,b)=\frac1{2}\beta\bigg(\frac{a+1}2,\frac{b+1}2\bigg)$$ Furthermore, we know that $$\beta(a,b)=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$ Which of course gives our integral: $$I(a,b)=\frac{\Gamma(\frac{a+1}2)\Gamma(\frac{b+1}2)}{2\Gamma(\frac{a+b}{2}+1)}$$ Plug in your values for $a,b$, and you'll be good to go.