# 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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• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – RE60K, heropup, Claude Leibovici, Adam Hughes, Najib Idrissi
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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

## 1 Answer

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