I came across this integral online
$$\int_{-\pi}^{\pi} \frac{x^3 \sin x}{1+e^{x \cos x}} dx$$
But every approach I've tried so far doesn't get me anywhere. I tried substituting $u=\cos x$, which doesn't really change much. Following that, I tried to separate the integrand with varying powers of $x$ in order to attempt an integration by parts, to no avail.
Any hints would be greatly appreciated.
Best Answer
We know, $$\int_a^b f(x) dx=\int_a^b f(a+b-x)dx$$ So $$I=\int_{-\pi}^\pi \frac{x ^3\sin x}{1+e^{x\cos x}}dx= \int_{-\pi}^\pi \frac{x ^3\sin x}{1+e^{-x\cos x}}dx $$$$\implies I=\int_{-\pi}^\pi \frac{e^{x\cos x}x ^3\sin x}{1+e^{x\cos x}}dx$$ Adding both the expressions, we get $$2I=\int_{-\pi}^\pi x^3\sin xdx=2\int_0^\pi x^3\sin xdx$$ (the integrand is even) so that $$I=\int_0^\pi x^3\sin xdx$$ and I hope you can take it from here.