-1
$\begingroup$

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

$\endgroup$

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "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
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ what did you tried, what's the context? $\endgroup$ – RE60K Sep 6 '14 at 14:36
  • $\begingroup$ Mathematica is unable to evaluate this integral. $\endgroup$ – Joshua Mundinger Sep 6 '14 at 14:37
7
$\begingroup$

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

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.