<s>[INST] Improper Integral $\int\limits_0^{\frac12}(2x - 1)^6\log^2(2\sin\pi x)\,dx$ How can I find a closed form for the following integral 
$$\int_0^{\frac12}(2x - 1)^6\log^2(2\sin\pi x)\,dx?$$
 [/INST] I tried the partial integration, with 
$\frac{d \log^2(2\sin(\pi x))}{dx} = \frac{1}{3}\log^3(2\sin(\pi x))\frac{1}{\frac{1}{2\sin(\pi x)}2cos(\pi x)\pi} = \frac{1}{3}\log^3(2\sin(\pi x))\frac{sin(\pi x)}{\pi cos(\pi x)} = \frac{\log^3(2\sin(\pi x))sin(\pi x)}{3\pi cos(\pi x)}$
$\int_0^{\frac{1}{2}}(2x-1)^2\log^2(2\sin(\pi x))\,dx =$
$[(2x-1)^2\frac{\log^3(2\sin(\pi x))sin(\pi x)}{3\pi cos(\pi x)}]_0^{\frac{1}{2}} - \int_0^{\frac{1}{2}}(4x-2)\log^2(2\sin(\pi x))\,dx =$
$[(2x-1)^2\frac{\log^3(2\sin(\pi x))sin(\pi x)}{3\pi cos(\pi x)}]_0^{\frac{1}{2}} + [(2 - 4x)\frac{\log^3(2\sin(\pi x))sin(\pi x)}{3\pi cos(\pi x)}]_0^{\frac{1}{2}} + 4\int_0^{\frac{1}{2}}\log^2(2\sin(\pi x))\,dx =$
$[(2x-1)^2\frac{\log^3(2\sin(\pi x))sin(\pi x)}{3\pi cos(\pi x)}]_0^{\frac{1}{2}} + [(2 - 4x)\frac{\log^3(2\sin(\pi x))sin(\pi x)}{3\pi cos(\pi x)}]_0^{\frac{1}{2}} + 4[\frac{\log^3(2\sin(\pi x))sin(\pi x)}{3\pi cos(\pi x)}]_0^{\frac{1}{2}}$


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*of course it's possible I messed it up somewhere

*with $\cos(\frac{\pi}{2})$ you have a $0$ in the denominator which makes it rather hard to evaluate the integral.

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