black integration

$\begin{aligned} I & = \int_0^\frac \pi 2 \sqrt[3]{\tan x} dx \\ & = \int_0^\frac \pi 2 \sin^\frac 13 x \cos^{-\frac 13} x \ dx & \small \blue{\text{Beta function }\text B(m,n) = 2 \int_0^\frac \pi 2 \sin^{2m-1} x \cos^{2n-1} x\ dx} \\ & = \frac 12 \text B \left(\frac 23, \frac 13 \right) & \small \blue{\text{and }\text B(m,n) = \frac {\Gamma(m)\Gamma(n)}{\Gamma(m+n)}\text{, }\Gamma(\cdot)\text{ is gamma function.}} \\ & \frac \blue{\Gamma\left(\frac 23\right)\Gamma\left(\frac 13\right)}{2\red{\Gamma(1)}} & \small \blue{\text{Note that }\Gamma(x)\Gamma(1-x) = \frac \pi {\sin (\pi x)}} \\ & = \frac \blue{\frac {2\pi}{\sqrt 3}}{2\cdot \red{0!}} = \frac \pi{\sqrt 3} \approx \boxed{1.81} & \small \red{\text{and } \Gamma (n) = (n-1)!} \end{aligned}$

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References:

• Beta function
• Gamma function

With the substitutions $y = \tan x$ and $z =y^2$ we obtain $J \; = \; \int_0^{\frac12\pi} \sqrt[3]{\tan x}\,dx \; = \; \int_0^\infty \frac{y^{\frac13}}{1 + y^2}\,dy \; = \; \tfrac12\int_0^\infty \frac{z^{-\frac13}}{1 + z}\,dz$ so that $J \; = \; \tfrac12B(\tfrac23,\tfrac13) \; =\; \frac{\pi}{2\sin(\frac13\pi)} \; =\; \boxed{\frac{\pi}{\sqrt{3}}}$

From the identity ( proof )

$\int_{0}^{\frac {\pi}{2}} \sqrt [n] {\tan x} dx = \frac {\pi}{2} \sec \left(\frac {\pi}{2n}\right)$

The answer is $\frac {\pi}{2} \cdot \sec \left(\frac {\pi}{6}\right) \approx \boxed {1.8137}$

So easy to solve.