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

Question

Problem: $$\int_{0}^{\frac{\pi}{2}} \frac{(\sec x)^{\frac{1}{3}}}{(\sec x)^{\frac{1}{3}}+(\tan x)^{\frac{1}{3}}} dx$$

My attempt:

I tried applying the property: $\int_{0}^{a} f(x)dx$ = $\int_{0}^{a} f(a-x)dx$ but got nowhere since the denominator changes. Even on adding the two integrals by taking LCM of the denominators, the final expression got more complicated because the numerator and denominator did not have any common factor.

I also tried dividing numerator and denominator by $(secx)^{\frac{1}{3}}$ to get $$\int_{0}^{\frac{\pi}{2}} \frac{1}{1+(\sin x)^{\frac{1}{3}}} dx$$ and then tried substituting $sinx$ = $t^3$ to get a complicated integral in $t$, which I couldn't evaluate.

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How do you evaluate this integral? (PS: If possible, please evaluate this without using special functions since this is a practice question for an entrance exam and we've only learnt some basic special functions and the gamma function.)

Answer 1

$$ \int_{0}^{\pi/2}\frac{1}{1+(\sin x)^{1/3}} = \int_{0}^{\pi/2}\frac{1-(\sin x)^{1/3}+(\sin x)^{2/3}}{1+\sin x}\,dx=I_1-I_2+I_3$$ 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:

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

Answer 2

Mathematica gives $$ \int_{0}^{\frac{\pi}{2}} \frac{(\sec x)^{\frac{1}{3}}}{(\sec x)^{\frac{1}{3}}+(\tan x)^{\frac{1}{3}}} dx= 1+\sqrt{\pi } \left(\frac{\Gamma \left(\frac{2}{3}\right)}{\Gamma \left(\frac{1}{6}\right)}-\frac{\Gamma \left(\frac{5}{6}\right)}{\Gamma \left(\frac{1}{3}\right)}+\frac{\Gamma \left(\frac{4}{3}\right)}{\Gamma \left(\frac{5}{6}\right)}-\frac{\Gamma \left(\frac{7}{6}\right)}{\Gamma \left(\frac{2}{3}\right)}\right)$$

I would try to integrate the power series $$\frac{(\sec x)^{\frac{1}{3}}}{(\sec x)^{\frac{1}{3}}+(\tan x)^{\frac{1}{3}}} \approx x^{2/3}+x^{4/3}-x^{5/3}-\frac{17 x^{7/3}}{18}+\frac{8 x^{8/3}}{9}-\frac{5 x^3}{6}+x^2-\sqrt[3]{x}-x+1 + O(x^{10/3})$$ (this is what mathematica gives, but it is worth to see the general term instead).

However, I'm afraid this wouldn't help to get that nice closed form, not straight forward.