# Show that $\int_0^{\pi /2}\frac{dx}{\sqrt{\cos{x}}}=\int_0^{\pi /2}\frac{dx}{\sqrt{\sin{x}}}=\frac{\Gamma^2(\frac{1}{4})}{2\sqrt{2 \pi}}$ [duplicate]

I am interested in showing that: $$\int_0^{\pi /2}\frac{dx}{\sqrt{\cos{x}}}=\int_0^{\pi /2}\frac{dx}{\sqrt{\sin{x}}}=\frac{\Gamma^2(\frac{1}{4})}{2\sqrt{2 \pi}}$$ I've come across this equality without anything else, would you please help me what I should do?

## marked as duplicate by Robert Z, Community♦Jul 18 '18 at 5:36

• Could you explain how? I'm quite unsure how to transform $\int_0^\infty t^{x-1}e^{-t}dt$ into $\int_0^{\pi /2}\frac{dx}{\sqrt{\sin{x}}}$... – Michal Dvořák Jul 18 '18 at 5:13
Using the substitution $u=\cos x$ gives $$\int_0^{\pi/2}\frac{dx}{\sqrt{\cos x}} =\int_0^1\frac{du}{u^{1/2}(1-u^2)^{1/2}}.$$ Now let $t=u^2$ to get $$\int_0^1\frac{du}{u^{1/2}(1-u^2)^{1/2}} =\frac12\int_0^1\frac{dt}{t^{3/4}(1-t)^{1/2}}=\frac12B(1/4,1/2)$$ where $B(x,y)$ is the Beta function. In terms of the Gamma function, $$\frac12B(1/4,1/2)=\frac{\Gamma(1/4)\Gamma(1/2)}{2\Gamma(3/4)} =\frac{\sqrt\pi\Gamma(1/4)}{2\Gamma(3/4)}.$$ From the identity $$\Gamma(x)\Gamma(1-x)=\frac\pi{\sin\pi x}$$ we get $$\Gamma(1/4)\Gamma(3/4)=\pi\sqrt2$$ and then $$\frac{\sqrt\pi\Gamma(1/4)}{2\Gamma(3/4)} =\frac{\Gamma(1/4)^2}{2\sqrt{2\pi}}.$$