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

Question

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? $$\int_{-a}^a \frac{1}{\cos^2 x (e^x + 1)} dx$$

Answer 1

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