I am working on an assignment question for my Advanced Calculus course and am having great difficulty working it out. In order to try and understand this type of question/working, I have found a similar question in one of my textbooks which, after attempting, I am having a similar issue with. The textbook provides a solution to problem, which I am struggling to understand. The question, solution and problem is as follows.
QUESTION: Show that $\int_0^2x\sqrt[3]{8-x^3}dx$ = $\frac{16\pi}{9\sqrt{3}}$
SOLUTION: After a few lines of changing integration variables, we are able to turn the integral into a Gamma function problem as follows:
LHS = $\frac{8}{9}\Gamma(\frac{1}{3})\Gamma(\frac{2}{3})$
Now, I completely understand all the working leading up to this point. I easily reached this point myself before referring to the solutions. The solution finishes as follows:
LHS = $\frac{8}{9}\Gamma(\frac{1}{3})\Gamma(\frac{2}{3}) = \frac{8}{9}\frac{\pi}{\sin{\pi/3}} = \frac{16\pi}{9\sqrt{3}}$ = RHS
I find it incredibly frustrating that the textbook solutions spell out almost every step in the working except for the only part that doesn't seem straightforward! I've been searching through the textbook and Google, trying to find out how to solve the gamma functions, but have not found anything helpful.
PROBLEM: How do we get from the $\Gamma(\frac{1}{3})\Gamma(\frac{2}{3})$ to $\frac{\pi}{\sin{\pi/3}}$?
Using Euler's reflection formula : $$\Gamma(z)\,\Gamma(1-z)=\frac {\pi}{\sin(\pi z)},\quad z\not\in\mathbb{Z}$$
For proofs see at MO starting with the Wiki proof.