How to prove that $\int_{-\pi}^{2\pi} f(x) dx \leq 0$ $$f(x) = \ln(x^2+1)\cdot e^{\sin(\lvert x \rvert)}\cdot\sin(x)$$
I need to prove that:
$$\int_{-\pi}^{2\pi} f(x) dx \leq 0$$
what I know is that $f$ is odd, thus $\int_{-\pi}^{\pi} f(x) dx = 0$ , so I'm left with proving that 
$$\int_{\pi}^{2\pi} f(x) dx \leq 0$$
Since $\int_{-\pi}^{2\pi} f(x) dx = \int_{-\pi}^{\pi} f(x) dx + \int_{\pi}^{2\pi} f(x) dx = 0 + \int_{\pi}^{2\pi} f(x) dx = \int_{\pi}^{2\pi} f(x) dx$
How do I continue from here?

$\sin x \leq 0$ for $x∈[π,2π]$, and all other factors are positive.
