How to calculate $\int_0^{\pi/2} \sqrt[3]{\mathrm{cotg}\; x} \;\mathrm{d}x$ I started by writting down: $$\mathrm{cotg} \;x=\frac{\cos x}{\sin x}$$ Furthermore, I have no idea.
 A: Let $\displaystyle I = \int_{0}^{\frac{\pi}{2}}(\cot x)^{\frac{1}{3}}dx\;,$ Now Put $\cot x= t^3\;,$ Then $\displaystyle -\csc^2 xdx = 3t^2dt\Rightarrow dx = -\frac{3t^2}{1+t^6}dt$
and changing Limits, We get $$I=-3\int_{0}^{\infty}\frac{t^3}{1+t^6}dt\;,$$ Now put $t^2=u\;,$ Then $\displaystyle 2tdt=du\Rightarrow tdt=\frac{du}{2} $
So we get $$I=-\frac{3}{2}\int_{0}^{\infty}\frac{u}{1+u^3}du = -\frac{3}{2}\int_{0}^{\infty}\frac{u}{(u+1)(u^2-u+1)}du$$
Now Using partial fraction.
$$\frac{u}{(u+1)(u^2-u+1)} = \frac{A}{u+1}+\frac{Bu+C}{u^2-u+1}$$
I have problems here. What should I do next ?
A: Hint, use a change of variable
$u=\mathrm{cotg} \;x$ to get
$$I=\int_0^{\pi/2} \sqrt[3]{\mathrm{cotg}\; x} \;\mathrm{d}x=\int_0^\infty \frac{\sqrt[3]{u}}{1+u^2}\mathrm{d}u$$
Then, another change of variable $u=v^3$:
$$I=\int_0^\infty\frac{3v^3}{1+v^6}\mathrm{d}v$$
Now you have an algebraic function to integrate, and the roots of the denominator are easy.
A: This is basically a Wallis integral, whose relation to the beta function is well-known, and can be shown by a simple trigonometric substitution. Then all that's left to do is using Euler's reflection formula for the $\Gamma$ function to simplify the expression, and finally arrive at the desired result. As for integrals of the form $\displaystyle\int_0^\infty\dfrac{x^{k-1}}{1+x^n}~dx,~$ a simple substitution $t=\dfrac1{1+x^n}$ will do the trick, yielding $~\dfrac\pi n~\csc\bigg(k~\dfrac\pi n\bigg)$ with the help of the same mechanism.
