The value of \( \int_{-\pi/3}^{\pi/3}\displaystyle\frac{1}{1+e^{\tan x}}dx \) is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \pi/3 \)
By the property \( \int_{a}^{b}f(x)\,dx=\int_{a}^{b}f(a+b-x)\,dx \) with a + b = 0, x can be replaced by −x, and \( \tan(-x)=-\tan x \):
\( \displaystyle I=\int_{-\pi/3}^{\pi/3}\frac{1}{1+e^{\tan x}}\,dx \)
\( \displaystyle =\int_{-\pi/3}^{\pi/3}\frac{1}{1+e^{-\tan x}}\,dx \)
\( \displaystyle =\int_{-\pi/3}^{\pi/3}\frac{e^{\tan x}}{e^{\tan x}+1}\,dx \)
Adding the two forms of I:
\( \displaystyle 2I=\int_{-\pi/3}^{\pi/3}\frac{1+e^{\tan x}}{1+e^{\tan x}}\,dx \)
\( \displaystyle =\int_{-\pi/3}^{\pi/3}1\,dx \)
\( \displaystyle =\frac{2\pi}{3} \)
\( \displaystyle I=\frac{\pi}{3} \)
Why the other options are incorrect
Option (2) \( -\pi/3 \) — Incorrect. The integrand is always positive, so the integral cannot be negative.
Option (3) \( \displaystyle\frac{2\pi}{3} \) — Incorrect. \( \frac{2\pi}{3} \) is 2I, the value before dividing by 2.
Option (4) \( \displaystyle\frac{\pi}{6} \) — Incorrect. \( \frac{\pi}{6} \) is half of the correct value; I equals half the length of the interval, \( \frac{1}{2}\cdot\frac{2\pi}{3} \), not a quarter of it.
Tip: Whenever f(x) + f(−x) = 1 on [−a, a], the integral is simply a, half the length of the interval.