I recently stumbled upon a question $$\int_0^{\infty}\frac{x^{m-1}\log^{a}x}{1+x^n}dx$$ I was able to evaluate it,but I am curious if there exists a closed form for, $$\int_0^{\pi/2}\frac{x^{2}\log{\sin x}}{1+x^6}dx$$ It numerically evaluates to -0.1392432458.
My attempt-
$$\int \frac {x^2}{1+x^6}dx=\frac13 \int \frac {d(x^3)}{1+x^6}=\frac13 \arctan {x^3}$$ Then,by applying integration by parts, $$\int_0^{\pi/2}\frac{x^2\log\sin x}{1+x^6}dx=-\int_0^{\pi/2}\frac13\arctan {x^3} \cot x dx$$.But now I'm stuck.