Show $$ \int_{-\pi}^{\pi} (4\arctan (e^x)-\pi)dx=0$$ without calculating.

I thought we should show that the integrand is odd, but I'm having trouble showing it.

If $f(x)=4\arctan (e^x)-\pi$ we need to show that $f(-x)=-f(x)$

so $f(-x)=4\arctan (e^{-x})-\pi$ and I'm not really see how to continue...

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    $\begingroup$ $\arctan(x) + \arctan(\frac1{x}) = \frac{\pi}{2}$ $\endgroup$ – Nicholas Nov 12 '15 at 11:33
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    $\begingroup$ @Nicholas exactly what I was missing, thanks! Do you want to make it an answer? $\endgroup$ – Stabilo Nov 12 '15 at 11:36

Note that $\arctan(x) + \arctan(\frac1{x}) = \frac{\pi}{2}$

Which can be derived by considering a right angle triangle.


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