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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.


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.)

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  • $\begingroup$ have you tried WolframAlpha to get an idea? $\endgroup$ – Dr. Sonnhard Graubner Jan 6 '16 at 18:46
  • $\begingroup$ @Dr.SonnhardGraubner Yeah. I got the numerical value of the integral as 0.872208 and the graph of the equation, but that doesn't help me in proceeding with the question. $\endgroup$ – Ashish Gupta Jan 6 '16 at 18:50
  • $\begingroup$ May I ask where you encountered this integral? (I don't see how to calculate it at the moment.) $\endgroup$ – mickep Jan 6 '16 at 19:00
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    $\begingroup$ @mickep My friend sent it to me as a doubt from one of his Maths books/assignments. I'll ask him the name of the book (if it is from one) and get back to you. $\endgroup$ – Ashish Gupta Jan 6 '16 at 19:05
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$$ \int_{0}^{\pi/2}\frac{1}{1+(\sin x)^{1/3}} = \int_{0}^{\pi/2}\frac{1-(\sin x)^{1/3}+(\sin x)^{2/3}}{1+\sin x}\,dx=I_1-I_2+I_3$$ 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:

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

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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.

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  • $\begingroup$ Thank you for providing this answer. However, I'm looking for a pencil-and-paper method to solve the question since it's for an entrance exam. I've edited in that detail in the question as well. $\endgroup$ – Ashish Gupta Jan 6 '16 at 19:20
  • $\begingroup$ @AshishGupta: I doubt that is the real statement of the original problem, probably there was a $\cot$ where $\sec$ is written. The answer to the question you gave above strongly depends on the values of the $\Gamma$ function at $\frac{1}{3}$ and $\frac{2}{3}$. $\endgroup$ – Jack D'Aurizio Jan 6 '16 at 19:31
  • $\begingroup$ @JackD'Aurizio You can check the question here. It's possible that there was a mistake in typing the question when the said page was written, but I haven't made any mistake in copying the question from my side. Although, looking at the level of the other questions, it does seem like there may have been a mistake in typing the question. $\endgroup$ – Ashish Gupta Jan 6 '16 at 19:41
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    $\begingroup$ @AshishGupta: I agree, there must have been a typo by the test composer, that made the exercise much more difficult. Anyway, it may be a good moment to learn something about the Euler beta function. It may be useful in the future. $\endgroup$ – Jack D'Aurizio Jan 6 '16 at 19:49

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