# How to integrate $\frac{x^{2}\log {\sin x}}{1+x^{6}}$

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.

• Where did you stumble upon the integral - what context? That sort of information would make this a better post. – Carl Mummert Mar 6 '15 at 11:41
• The question I stumbled upon had $\log x$ instead of $\log \sin x$. – Akshay Bodhare Mar 6 '15 at 12:42