Integral related to $\sum\limits_{n=1}^\infty\sin^n(x)\cos^n(x)$ 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?

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...
