<s>[INST] How to solve the integration How to find $$ \frac{\int_0^{\pi}x^3\log(\sin x)\,dx}{\int_0^{\pi} x^2 \log(\sqrt{2} \sin x)\,dx} $$
I couldn’t resolve it by using integration by parts.
 [/INST] Using Fourier series, we see that
\begin{align}
\log(\sin x)= -\log 2-\sum^\infty_{k=1}\frac{\cos(2kx)}{k}
\end{align}
for $0\le x\le \pi$. Hence we have
\begin{align}
\int^\pi_0x^3\log(\sin x)\ dx =&\ -\log 2\int^\pi_0x^3\ dx -\sum^\infty_{k=1} \frac{1}{k}\int^\pi_0 x^3\cos(2kx)\ dx\\
=&\ -\frac{\pi^4}{4}\log 2 - \frac{3\pi^2}{4}\sum^\infty_{k=1} \frac{1}{k^3} = -\frac{\pi^4}{4}\log 2- \frac{3\pi^2}{4}\zeta(3).
\end{align}
On the other hand, we have
\begin{align}
\int^\pi_0 x^2\log(\sqrt{2}\sin x)\ dx =&\  \int^\pi_0 \frac{1}{2}x^2\log 2+ x^2\log(\sin x)\ dx\\
=&\ \frac{\pi^3}{6}\log 2 + \int^\pi_0 x^2\log(\sin x)\ dx\\
=&\ \frac{\pi^3}{6}\log 2 -\log 2\int^\pi_0x^2\ dx -\sum^\infty_{k=1} \frac{1}{k}\int^\pi_0 x^2\cos(2kx)\ dx\\
=&\ -\frac{\pi^3}{6}\log 2 -\frac{\pi}{2}\sum^\infty_{k=1} \frac{1}{k^3}= -\frac{\pi^3}{6}\log 2 - \frac{\pi}{2}\zeta(3). 
\end{align}
Then we see that
\begin{align}
\frac{\int^\pi_0 x^3 \log(\sin x)\ dx}{\int^\pi_0 x^2\log(\sqrt{2}\sin x)\ dx} = \frac{-\frac{\pi^4}{4}\log 2- \frac{3\pi^2}{4}\zeta(3)}{-\frac{\pi^3}{6}\log 2 - \frac{\pi}{2}\zeta(3)} = \frac{3\pi}{2}.
\end{align}
</s>