How to solve the following integral $\int_0^{\frac{\pi}{2}}\sqrt[3]{\sin^8x\cos^4x}dx$? How to solve the following integral?
$$\int_0^{\frac{\pi}{2}}\sqrt[3]{\sin^8x\cos^4x}\,dx$$
Preferably without the universal substitution $$\sin(t) = \dfrac{2\tan(t/2)}{1+\tan^2(t/2)}$$
 A: Using $\operatorname{B}(a,\,b)=2\int_0^{\pi/2}\sin^{2a-1}x\cos^{2b-1}xdx$, your integral is$$\frac12\operatorname{B}\left(\frac{11}{6},\,\frac{7}{6}\right)=\frac{\Gamma\left(\frac{11}{6}\right)\Gamma\left(\frac{7}{6}\right)}{2\Gamma(3)}=\frac{5}{144}\Gamma\left(\frac{5}{6}\right)\Gamma\left(\frac{1}{6}\right)=\frac{5\pi}{144}\csc\frac{\pi}{6}=\frac{5\pi}{72}.$$Here the first $=$ uses $\operatorname{B}(a,\,b)=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$, the second $\Gamma(a+1)=a\Gamma(a)$, the third $\Gamma(a)\Gamma(1-a)=\pi\csc\pi a$.
A: @J.G. Mentioned the use of $$\int_0^{\pi/2}\cos(x)^{2a-1}\sin(x)^{2b-1}dx=\frac{\Gamma(a)\Gamma(b)}{2\Gamma(a+b)},$$
which I will prove for you here.
Recall the definition of the Gamma function:
$$\Gamma(s)=\int_0^\infty t^{s-1}e^{-t}dt\qquad \rm{Re }(s)>0.$$
Setting $t=x^2$ gives 
$$\Gamma(s)=2\int_0^\infty x^{2s-1}e^{-x^2}dx.$$
Thus,
$$\Gamma(a)\Gamma(b)=4\int_0^\infty \int_0^\infty x^{2a-1}y^{2b-1}e^{-(x^2+y^2)}dxdy.$$
Then we convert the integrals to polar coordinates to get
$$\begin{align}
\Gamma(a)\Gamma(b)&=4\int_0^{\pi/2}\int_0^{\infty} r(r\cos\theta)^{2a-1}(r\sin\theta)^{2b-1}e^{-r^2}drd\theta\\
&=4\int_0^{\pi/2}\cos(\theta)^{2a-1}\sin(\theta)^{2b-1}\int_0^{\infty} r^{2a+2b-1}e^{-r^2}drd\theta\\
&=2\left(2\int_0^{\infty} r^{2a+2b-1}e^{-r^2}dr\right)\left(\int_0^{\pi/2}\cos(\theta)^{2a-1}\sin(\theta)^{2b-1}d\theta\right)\\
&=2\Gamma(a+b)\int_0^{\pi/2}\cos(\theta)^{2a-1}\sin(\theta)^{2b-1}d\theta .
\end{align}$$
You could even take this a step further and set $t=\cos^2\theta$ to get the original Beta integral:
$$\int_0^1 t^{a-1}(1-t)^{b-1}dt=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)},$$
from which many other results can be derived. 
