Finding $\displaystyle \int^{\pi}_{0}x^2\ln(\sin x)dx$
Solution I try
Assume $\displaystyle I =\int^{\pi}_{0}x^2\ln(\sin x)dx=\int^{\pi}_{0}x^2\ln(2\sin x/2 \cos x/2)dx$
$\displaystyle I=\int^{\pi}_{0}x^2\ln(\sin \frac{x}{2})dx+\int^{\pi}_{0}x^2\ln(\cos \frac{x}{2})dx+\int^{\pi}_{0}x^2\ln(2)dx$
$\displaystyle I=8\int^{\frac{\pi}{2}}_{0}t^2\ln\sin(t)dt+8\int^{\frac{\pi}{2}}_{0}t^2\ln\cos(t)dt+2\int^{\frac{\pi}{2}}_{0}\ln(2)dt$
I do not understand how can I solve it.
Hint Begin by $$ \displaystyle \int^{v}_{u}x^2\ln(\sin x)dx=\displaystyle \int^{v}_{u}\Big(\frac{x^3}{3}\Big)'\ln(\sin x)dx $$ and use integration by parts.
Of course, you will have to prove the convergence when $u\to 0$ and $v\to \pi$.
You can also use the Fourier series for $0<x<\pi$
$$
-\log(\sin(x))=\sum_{k=1}^\infty\frac{\cos(2kx)}{k}+\log(2)
$$
If needed, look at the very elegant derivation of it here by user17762.
And then use, as was indicated, the technique of Jacky Chong
For your information, computer algebra gives (same as Dr. Sonnhard Graubner) $$ -1/6\pi^3\log(4) - 1/2\pi\zeta(3)\ . $$
