integration - How to solve the following integral $int_0^{frac{pi}{2}}sqrt[3]{sin^8xcos^4x}dx$?

How to solve the following integral?

$$\int_0^{\frac{\pi}{2}}\sqrt[3]{\sin^8x\cos^4x}\,dx$$

Preferably without the universal substitution $$\sin(t) = \dfrac{2\tan(t/2)}{1+\tan^2(t/2)}$$

Answer

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 $\Gamma(a)\Gamma(1-a)=\pi\csc\pi a$.