What is the value of the ratio $$\frac{\int_{0}^{\pi}x^3\log(\sin x) dx}{\int_{0}^{\pi}x^2\log(\sqrt{2}\sin x)dx}$$ ?
Can someone suggest me a nice and quick approach to this problem?
I actually integrated the numerator but then i'm feeling that traditional approach is too long!!Help me out.
There would be one such approach. Let $J_k=\displaystyle\int_0^\tfrac\pi2x^k\ln\sin x~dx.~$ Then, dividing $(0,~\pi)$ into
$\bigg(0,\dfrac\pi2\bigg)$ and $\bigg(\dfrac\pi2,~\pi\bigg),~$ and substituting $x\mapsto\pi-x$ on the latter $\big($along with using $\ln ab=$
$=\ln a+\ln b$ in the denominator$\big)$, we have $F=\pi\cdot\dfrac{\pi^2J_0-3\pi J_1+3J_2}{\dfrac{\pi^3\ln2}6+\pi^2J_0-2\pi J_1+2J_2}~.~$ Now,
as it happens, $J_0=-\dfrac{\pi\ln2}2~.~$ Thus, without having to evaluate either $J_1$ or $J_2,~$ we can
easily see that $F=\dfrac32\cdot\pi.~$ But evaluating $J_0$ is trivial, by substituting $x\to\dfrac\pi2-x,~$ adding
the two equivalent expressions together, using $\ln a+\ln b=\ln ab$ in conjunction with $\sin2t=$
$=2\sin t~\cos t,~$ to ultimately arrive at $2J_0=J_0-\displaystyle\int_0^\tfrac\pi2\ln2~dx,~$ from which the much desired
conclusion inevitably follows.
