# Evaluate the integral $\int_{-1}^{1}\frac{\sin x}{6+x^2}\,dx$

I need to calculate the following integral: $$\int_{-1}^{1} \frac{\sin x}{6+x^2} dx$$ I have tried trigonometrical substitution like $$x=a\tan \theta$$ and so on but i cannot solve it ... any ideas?

Thank you very much!

If $f$ is an odd function integrable on $[-a,a]$ then $$\int_{-a}^{a}f(x)\,dx=0$$
the result is zero since the function $$f(x)=\frac{\sin(x)}{6+x^2}$$ is odd.