Question:

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

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\(\sqrt{\tan^2x}=|\tan x|\). Split the interval at \(2\pi\), where \(\tan x\) changes sign from negative to positive.
Updated On: Oct 31, 2026
  • \(\log 2\sqrt{2}\)
  • \(\log 2\)
  • \(2\log 2\)
  • \(\log\sqrt{2}\)
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The Correct Option is A

Solution and Explanation

\(\sqrt{\tan^2x}=|\tan x|\). On \((7\pi/4,\,2\pi)\) (fourth quadrant) \(\tan x \lt 0\); on \((2\pi,\,7\pi/3)\) \(\tan x \gt 0\).
\(I=\int_{7\pi/4}^{2\pi}(-\tan x)\,dx+\int_{2\pi}^{7\pi/3}\tan x\,dx=\Big[\ln|\cos x|\Big]_{7\pi/4}^{2\pi}+\Big[-\ln|\cos x|\Big]_{2\pi}^{7\pi/3}\).
First part: \(\ln1-\ln\frac{1}{\sqrt2}=\ln\sqrt2\).
Second part: \(-\ln\frac12+\ln1=\ln2\).
\(I=\ln\sqrt2+\ln2=\ln(2\sqrt2)\).
Answer: (1) \(\log 2\sqrt{2}\)
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