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.
The integration you want to solve doesn't have an analytical solution, so using integration by parts won't help you much. However, you can use a number of numerical methods out there for approximating integrals.
The integration you want to solve doesn't have an analytical solution, so using integration by parts won't help you much. However, you can use a number of numerical methods out there for approximating integrals.