On a tricky integral and Euler's reflection formula

Question

I've been trying to compute the following integral $\forall s\in]0;1[$: $$I=\int_0^{\frac{\pi}{2}}\cot^{2s-1}(\theta)\ d\theta$$ The limits smelled a lot like the Beta function, and sure enough, playing around yields the following result: $$I=\int_0^{\frac{\pi}{2}}\tan(\theta)\cot^{2s}(\theta)\ d\theta$$ $$=\int_0^{\frac{\pi}{2}}\cos^{2s-1}(\theta)\sin^{1-2s}(\theta)\ d\theta$$ $$=\frac{1}{2}\beta(s,1-s)$$ Where of course $\beta(m,n)$ is the Eulerian integral of the first kind. This integral is conseqeuntly equal to: $$I=\frac{1}{2}\Gamma(s)\Gamma(1-s)$$ And via Euler's reflection formula: $$I=\frac{\pi}{2\sin(\pi s)}$$ My question is whether there exists another way to get to this result without ever having to go through the Gamma or Beta functions, whether we can sort of derive Euler's reflection formula without even having to use them, and just by studying this integral. I'm curious to hear your thoughts on this. As always, this is purely recreational!

Answer 1

If you like the Gaussian hypergeometric function $$I=\int \cot ^a(\theta )\,d\theta$$ $$I=-\frac{\cot ^{a+1}(\theta ) }{a+1}\, _2F_1\left(1,\frac{a+1}{2};\frac{a+3}{2};-\cot ^2(\theta )\right)$$

Assuming $a<1$ and $0<t < \frac \pi 2$ $$J=\int_0^t \cot ^a(\theta )\,d\theta$$ $$J=\frac{\pi}{2} \, \sec \left(\frac{\pi a}{2}\right)- \frac{\cot ^{a+1}(t) }{a+1}\, _2F_1\left(1,\frac{a+1}{2};\frac{a+3}{2};-\cot^2(t)\right)$$ and $$\underset{t\to \frac{\pi }{2}}{\text{limit}}\left(\,_2F_1\left(1,\frac{a+1}{2};\frac{a+3}{2};-\cot^2(t)\right)\right)=1$$