TITLE: Quotient of two integrals $\frac{\int_0^\pi x^3\ln(\sin x)\,dx}{\int_0^\pi x^2\ln(\sqrt{2}(\sin x))\,dx}$
QUESTION [5 upvotes]: Calculate
  $$\frac{\int_0^\pi x^3\ln(\sin x)\,dx}{\int_0^\pi x^2\ln(\sqrt{2}(\sin x))\,dx}$$

In this problem, I'm unable to understand how to start.
I tried applying integration by parts but I couldn't solve it. I also tried the various properties of definite integration but they were of no use.
Maybe applying integration by parts (or DI method) successively may work but it leads to a form of $\frac{\infty}{\infty}$.

REPLY [3 votes]: For those interest in the overkill approach, I will be providing closed forms for each integral with the use of special functions just for the hell of it.

We define
$$p=\int_0^\pi x^3\ln\sin x\,dx$$
We recall the definition of the Clausen function of order $2$:
$$\mathrm{Cl}_2(x)=-\int_0^x \ln\left|2\sin\frac{t}2\right|\,dt=\sum_{k\geq1}\frac{\sin kx}{k^2}$$
so
$$-\ln\left(2\sin \frac{x}2\right)=\mathrm{Cl}_1(x)=\sum_{k\geq1}\frac{\cos kx}{k}$$
and thus $$\ln\sin x=-\ln2-\sum_{k\geq1}\frac{\cos2kx}{k}$$
then
$$\begin{align}
p&=-\int_0^\pi x^3\left(\ln2+\sum_{k\geq1}\frac{\cos2kx}{k}\right)dx\\
&=-\frac{\pi^4}4\ln2-\frac1{16}\sum_{k\geq1}\frac1{k^5}\int_0^{2k\pi}x^3\cos x\,dx
\end{align}$$
We can use IBP to show that
$$\int_0^{2k\pi}x^3\cos x\,dx=12\pi^2k^2$$
Which I leave to you as a challenge.
Long story short,
$$p=-\frac{\pi^4}{4}\ln2-\frac{3\pi^2}4\zeta(3)$$
Where $\zeta(3)=\sum_{k\geq1}k^{-3}$ is Apery's Constant. And $\zeta(s)=\sum_{k\geq1}k^{-s}$ is the Riemann Zeta function.

Next up:
$$q=\int_0^\pi x^2\ln(\sqrt{2}\sin x)\,dx=\frac{\pi^3}{6}\ln2+\int_0^\pi x^2\ln\sin x\,dx$$
Using the same series as last time,
$$\begin{align}
\int_0^\pi x^2\ln\sin x\,dx&=-\frac{\pi^3}{3}\ln2-\frac18\sum_{k\geq1}\frac1{k^4}\int_0^{2k\pi}x^2\cos x\,dx
\end{align}$$
IBP shows that $$\int_0^{2k\pi}x^2\cos x\,dx=4\pi k$$
So of course
$$\int_0^\pi x^2\ln\sin x\,dx=-\frac{\pi^3}{3}\ln2-\frac\pi2\zeta(3)$$
Hence
$$q=-\frac{\pi^3}{6}\ln2-\frac\pi2\zeta(3)$$

So the ratio in question is 
$$\frac{p}{q}=\frac{\frac{\pi^4}{4}\ln2+\frac{3\pi^2}4\zeta(3)}{\frac{\pi^3}{6}\ln2+\frac\pi2\zeta(3)}=\frac32\pi$$