# How to evaluate $\int \frac{1}{\cos^2 x (e^x + 1)} \,dx$? [closed]

The indefinite integral $$\int \frac{1}{\cos^2 x (e^x + 1)} dx$$ appears to be impossible to evaluate in closed form.

Could you please suggest how I should evaluate this integral in definite form? $$\int_{-a}^a \frac{1}{\cos^2 x (e^x + 1)} dx$$

## closed as off-topic by RE60K, heropup, Claude Leibovici, Adam Hughes, Najib IdrissiMar 3 '15 at 8:13

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• what did you tried, what's the context? – RE60K Sep 6 '14 at 14:36
• Mathematica is unable to evaluate this integral. – Joshua Mundinger Sep 6 '14 at 14:37

I suppose you are integrating over an interval: $$\int_{-a}^{a} \frac{1}{\cos^2 x (e^x + 1)} dx=\int_0^a\frac{dx}{\cos^2 x(e^x+1)}+\frac{dx}{\cos^2x(1+e^{-x})}=\int_0^a\frac{dx}{\cos^2 x}=[\tan x]_0^a=\tan a$$