3

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

I tried much of elementary methods to solve above integral but is not advancing. Any methods from elementary to advanced are appreciated.

omegadot
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TerenceP
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    Using WolframAlpha to integrate over $[0, \pi]$ (and appealing to symmetry) gives a solution in terms of hypergeometric functions and the Gamma function, so there's probably not a closed form in terms of elementary functions. – Travis Willse Mar 05 '19 at 08:04
  • Based on Wolfram's evaluation of the antiderivative (not the definite integral), it's not going to be pretty (https://www.wolframalpha.com/input/?i=antiderivative+of+(sin(x))%5E(1%2F3)+%2F+(4+-+sin(x)%5E2)) and will involve the Appell hypergeometric bivariate function (http://mathworld.wolfram.com/AppellHypergeometricFunction.html). So while there is kind of a closed form, I feel like some sort of numerical approximation would be much more suitable unless you absolutely need an exact result for wherever this came up. – PrincessEev Mar 05 '19 at 08:06
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    answer is given $\frac{\pi}{2^{2/3} 3^{3/2}}$ I think some different form is attainable. – TerenceP Mar 05 '19 at 08:07

2 Answers2

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Hint. One may write $$ \frac{\sqrt[3]{\sin x}}{4-\sin^2 x}=\frac14\sum_{n=0}^\infty\frac1{4^n}\left(\sin x\right)^{2n+\frac13} $$ then one is allowed to perform a termwise integration $$ \int_0^{\Large \frac{\pi}2}\frac{\sqrt[3]{\sin x}}{4-\sin^2 x}\,dx=\sum_{n=0}^\infty\frac1{4^{n+1}}\int_0^{\Large \frac{\pi}2}\left(\sin x\right)^{2n+\frac13}\,dx $$using the classic Euler evaluation $$ \int_0^{\Large \frac{\pi}2}\left(\sin x\right)^s\,dx=\frac{\sqrt{\pi} \,\Gamma \left(\frac{s+1}{2}\right)}{2\, \Gamma \left(\frac{s}{2}+1\right)},\qquad s>0, $$ obtaining

$$ \int_0^{\Large \frac{\pi}2}\frac{\sqrt[3]{\sin x}}{4-\sin^2 x}\,dx=\frac{3 \sqrt{\pi} \,\, _2F_1\left(\frac{2}{3},1;\frac{7}{6};\frac{1}{4}\right)\, \Gamma \left(\frac{2}{3}\right)}{4\, \Gamma \left(\frac{1}{6}\right)}. $$

A path to the simplification $\dfrac{\pi}{2^{2/3} 3^{3/2}}$ would be interesting.

Olivier Oloa
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3

Here is a way to simplify @Olivier Oloa's answer for the integral that is written in terms of the Gauss hypergeometric function.

Starting from (4) from a previous answer I gave here, it was shown that $$_2F_1 \left (1, \frac{2}{3}; \frac{7}{6}; \frac{1}{4} \right ) = \frac{2^{4/3}}{\sqrt{\pi}} \Gamma \left (\frac{4}{3} \right ) \Gamma \left (\frac{7}{6} \right ).$$

Thus \begin{align} I &= \frac{3 \sqrt{\pi}}{2^2} \cdot \frac{2^{4/3}}{\sqrt{\pi}} \cdot \frac{\Gamma (\frac{4}{3}) \Gamma (\frac{7}{6}) \Gamma (\frac{2}{3})}{\Gamma (\frac{1}{6})}\\ &= \frac{1}{2^{2/3} 6} \Gamma \left (1 - \frac{1}{3} \right ) \Gamma \left (\frac{1}{3} \right )\\ &= \frac{1}{2^{2/3} 6} \cdot \frac{\pi}{\sin (\frac{\pi}{3})}\\ &= \frac{\pi}{2^{2/3} 3^{3/2}}, \end{align} giving the required simplification.

Actually, the reason for this coincidence is, as I have just discovered, unsurprising. As @Zacky correctly observes in the comments section of the linked question, the integral considered here is, to within a numerical factor, just that considered in the linked question, namely $$\int_0^1 \frac{x^{2/3}}{\sqrt[3]{1 - x} (1 - x + x^2)} \, dx = 2^{5/3} \int_0^{\frac{\pi}{2}} \frac{\sqrt[3]{\sin x}}{4 - \sin^2 x} \, dx.$$

omegadot
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