How do you evaluate $\int_{0}^{\frac{\pi}{2}} \frac{(\sec x)^{\frac{1}{3}}}{(\sec x)^{\frac{1}{3}}+(\tan x)^{\frac{1}{3}}} \, dx ?$

$$ \int_{0}^{\pi/2}\frac{1}{1+(\sin x)^{1/3}} = \int_{0}^{\pi/2}\frac{1-(\sin x)^{1/3}+(\sin x)^{2/3}}{1+\sin x}\,dx=I_1-I_2+I_3$$ where: $$ I_1 = \int_{0}^{\pi/2}\frac{dx}{1+\cos x}=\int_{0}^{\pi/2}\frac{1-\cos x}{\sin^2 x}\,dx = \left.\left(\csc x-\cot x\right)\right|_{0}^{\pi/2}=1,$$ $$ I_2 = \int_{0}^{\pi/2}\frac{(\cos x)^{1/3}-(\cos x)^{4/3}}{\sin^2 x}\,dx,\quad I_3 = \int_{0}^{\pi/2}\frac{(\cos x)^{2/3}-(\cos x)^{5/3}}{\sin^2 x}\,dx $$ but Euler's beta function gives: $$ \int_{0}^{\pi/2}(\sin x)^\alpha (\cos x)^{\beta}\,dx = \frac{\Gamma\left(\frac{\alpha+1}{2}\right)\cdot\Gamma\left(\frac{\beta+1}{2}\right)}{2\cdot\Gamma\left(\frac{2+\alpha+\beta}{2}\right)}$$ hence, after some simplification:

$$ \int_{0}^{\pi/2}\frac{dx}{1+(\sin x)^{1/3}} = 1-\frac{2^{4/3}\pi^2(\sqrt{3}-1)}{3\cdot\Gamma\left(\frac{1}{3}\right)^3}+\frac{2^{2/3}\pi^2(2-\sqrt{3})}{9\cdot\Gamma\left(\frac{2}{3}\right)^3}. $$