A definite integration problem

Question

$$\int_{-\pi}^\pi \frac{\cos^2x}{1+a^x}dx$$ where $a>0$

This question is from my textbook. I am finding this question quite difficult to solve.

Take advantage of a symmetry. – Mark Viola Oct 12 '21 at 15:06 — Mark Viola, Oct 12 '21 at 15:06
Use $\int_{-a}^{a} f(x) dx=\int_{0}^{a} [f(x)+f(-x)] dx$ – Z Ahmed Oct 12 '21 at 15:12 — Z Ahmed, Oct 12 '21 at 15:12

Mark Viola · Accepted Answer · 2021-10-12T15:11:52.420

3

HINT:

Note that we have

$$\int_{-\pi}^0 \frac{\cos^2(x)}{1+a^x}\,dx=\int_0^\pi \frac{a^x\cos^2(x)}{1+a^x}\,dx$$

Can you finish?

edited Oct 12 '21 at 15:11

answered Oct 12 '21 at 15:08

Mark Viola

Can you please elaborate on how that symmetry would occur? – Krish Oct 12 '21 at 15:11
Thank you, I just understood that symmetry. – Krish Oct 12 '21 at 15:13
Pleased to hear. – Mark Viola Oct 12 '21 at 15:16

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