Question

The integral \(\int_{7\pi/4}^{7\pi/3}\sqrt{\tan^2x}\,dx\) is equal to :

• A. \(\log 2\sqrt{2}\)
• B. \(\log 2\)
• C. \(2\log 2\)
• D. \(\log\sqrt{2}\)

Hint:

\(\sqrt{\tan^2x}=|\tan x|\). Split the interval at \(2\pi\), where \(\tan x\) changes sign from negative to positive.

Answer 1

The Correct Option is A

\(|\tan x|\) has period \(\pi\) (and \(2\pi\)), so shift the limits by \(-2\pi\): \(\int_{7\pi/4}^{7\pi/3}|\tan x|\,dx=\int_{-\pi/4}^{\pi/3}|\tan x|\,dx\).
Split at 0: \(\int_{-\pi/4}^{0}(-\tan x)\,dx+\int_0^{\pi/3}\tan x\,dx\).
By symmetry \(\int_{-\pi/4}^{0}(-\tan x)\,dx=\int_0^{\pi/4}\tan x\,dx=\ln\sec\frac{\pi}{4}=\ln\sqrt2\).
\(\int_0^{\pi/3}\tan x\,dx=\ln\sec\frac{\pi}{3}=\ln2\).
Total \(=\ln\sqrt2+\ln2=\ln2\sqrt2\) (about 1.04).
Answer: (1) \(\log 2\sqrt{2}\)