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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!

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Hint:

If $f$ is an odd function integrable on $[-a,a]$ then $$\int_{-a}^{a}f(x)\,dx=0$$

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the result is zero since the function $$f(x)=\frac{\sin(x)}{6+x^2}$$ is odd.

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