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Playing around in Mathematica, I found the following:

$$\int_0^\pi\sum_{n=1}^\infty\sin^n(x)\cos^n(x)\ dx=0.48600607\ldots =\Gamma(1/3)\Gamma(2/3)-\pi.$$

I'm curious... how could one derive this?

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    $\begingroup$ So, you're asking about the integral $$\int_0^\pi \frac{\sin\,2u}{2-\sin\,2u} \mathrm du$$ I suppose... $\endgroup$ – J. M. isn't a mathematician Apr 19 '12 at 10:19
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    $\begingroup$ Also: $\Gamma\left(\frac13\right)\Gamma\left(1-\frac13\right)=\frac{\pi}{\sin\frac{\pi}{3}}$ $\endgroup$ – J. M. isn't a mathematician Apr 19 '12 at 10:20
  • $\begingroup$ @J.M.: ...well, when you put it that way... :) I suppose I should study elementary calculus properly before going off on a wild-goose chase. Thanks! $\endgroup$ – Carolus Apr 19 '12 at 10:23
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I think you are making this much more difficult than it has to be. Since $|\sin(x)\cos(x)|< 1$ you have that

$$\sum_{n=1}^{\infty}(\sin(x)\cos(x))^n=\frac{\sin(x)\cos(x)}{1-\sin(x)\cos(x)}=\frac{\sin(2x)}{2-\sin(2x)}$$

And you can just find through normal calc that

$$\int \frac{\sin(2x)}{2-\sin(2x)}\mathrm dx=-\left(x+\frac{2}{\sqrt{3}}\arctan\left(\frac{1-2\tan(x)}{\sqrt{3}}\right)\right)$$

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  • $\begingroup$ Got it. Seems like I was posting before thinking... Thanks! $\endgroup$ – Carolus Apr 19 '12 at 10:24
  • $\begingroup$ @Alex I think you would be better off with $|\sin(x)\cos(x)|\leqslant 1/2$ or $|\sin(x)\cos(x)|<1$ , otherwise you would have a problem for those values which the equality holds. $\endgroup$ – Pedro Tamaroff Apr 20 '12 at 2:56
  • $\begingroup$ Of course, thank you. $\endgroup$ – Alex Youcis Apr 20 '12 at 2:57
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For giggles:

$$\begin{align*} \sum_{n=1}^\infty\int_0^\pi \sin^n u\,\cos^n u\;\mathrm du&=\sum_{n=1}^\infty\frac1{2^n}\int_0^\pi \sin^n 2u\;\mathrm du\\ &=\frac12\sum_{n=1}^\infty\frac1{2^n}\int_0^{2\pi} \sin^n u\;\mathrm du\\ &=\frac12\sum_{n=1}^\infty\frac1{2^{2n}}\int_0^{2\pi} \sin^{2n} u\;\mathrm du\\ &=2\sum_{n=1}^\infty\frac1{2^{2n}}\int_0^{\pi/2} \sin^{2n} u\;\mathrm du\\ &=\pi\sum_{n=1}^\infty\frac1{2^{4n}}\binom{2n}{n}=\pi\sum_{n=1}^\infty\frac{(-4)^n}{16^n}\binom{-1/2}{n}\\ &=\pi\left(\frac1{\sqrt{1-\frac14}}-1\right)=\pi\left(\frac2{\sqrt 3}-1\right) \end{align*}$$

where the oddness of the sine function was used in the third line to remove zero terms, the Wallis formula and the binomial identity $\dbinom{2n}{n}=(-4)^n\dbinom{-1/2}{n}$ were used in the fifth line, after which we finally recognize the binomial series and evaluate accordingly.

Of course, Alex's solution is vastly more compact...

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    $\begingroup$ Yay Wallis! :-) $\endgroup$ – Aryabhata Apr 19 '12 at 16:46
  • $\begingroup$ @Aryabhata: I wrote this answer with you in mind... :) I'm relieved you managed to see it. $\endgroup$ – J. M. isn't a mathematician Apr 19 '12 at 16:49
  • $\begingroup$ @J.M. Giggles are in order. This problem summarizes some of my favs! I will add it to my binder. $\endgroup$ – Pedro Tamaroff Apr 20 '12 at 2:18

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