Quotient of two integrals $\frac{\int_0^\pi x^3\ln(\sin x)\,dx}{\int_0^\pi x^2\ln(\sqrt{2}(\sin x))\,dx}$ 
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 $\frac{\infty}{\infty}$.
 A: 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 
$$\frac{p}{q}=\frac{\frac{\pi^4}{4}\ln2+\frac{3\pi^2}4\zeta(3)}{\frac{\pi^3}{6}\ln2+\frac\pi2\zeta(3)}=\frac32\pi$$
A: We want to prove that: $$\frac{I}{J}=\frac{\int_0^\pi x^3\ln(\sin x)dx} {\int_0^\pi x^2\ln\left(\sqrt 2\sin x\right)dx}=\frac{3\pi}2$$
Let's take the upper integral and substitute $\pi-x\to x$ and add a $0$ term in the end:
$$\Rightarrow I=\int_0^\pi (\pi^3-3\pi^2x+3\pi x^2-x^3)\ln(\sin x)dx+ 3\pi(\underbrace{\ln \sqrt 2-\ln \sqrt 2}_{=0})\int_0^\pi x^2 dx$$
$$\small=\pi^3 \int_0^\pi \ln(\sin x)dx-3\pi^2 \int_0^\pi x\ln(\sin x)dx+3\pi\int_0^\pi x^2(\ln(\sin x)+\ln\sqrt 2)dx-I-{\pi^4}\ln \sqrt 2$$
$$\small \Rightarrow 2I=\left(\pi^3-\frac{3\pi^3}{2}\right)\int_0^\pi \ln(\sin x)dx-{\pi^4}\ln \sqrt 2+3\pi\int_0^\pi x^2\ln(\sqrt2 \sin x)dx$$
$$\require{cancel} 2I=\cancel{\frac{\pi^3}{2}\cdot 2\pi \ln \sqrt 2}-\cancel{\pi^4 \ln \sqrt 2}+3\pi J\Rightarrow I=\frac{3\pi}2J$$

Things used above: $$K=\int_0^\pi x\ln(\sin x)dx=\int_0^\pi (\pi-x)\ln(\sin x)dx$$
$$\Rightarrow 2K=\int_0^\pi (x+\pi-x)\ln(\sin x)dx\Rightarrow K=\frac{\pi}{2}\int_0^\pi \ln(\sin x)dx$$
$$L=\int_0^\pi \ln(\sin x)dx=\int_0^\frac{\pi}{2} \ln(\sin x)dx+\int_0^\frac{\pi}{2} \ln(\cos x)dx$$
$$=\int_0^\pi \ln\left(\frac22\sin x\cos x\right)=\int_0^\frac{\pi}{2} \ln(\sin 2x)dx-\int_0^\frac{\pi}{2} \ln 2dx$$
$$=\frac12 \int_0^\pi \ln(\sin x) dx-\ln\sqrt 2 \int_0^{\pi} dx\Rightarrow L=-2\pi \ln\sqrt 2$$
