Question

The value of definite integral $\underset{\frac{7\pi }{4}}{\overset{\frac{7\pi }{3}}{\int }}\sqrt{{\tan }^{2}x}dx$ is equal to

• A. $\ln 2$
• B. $\ln 4$
• C. $\frac{1}{2}\ln 2$
• D. $\frac{3}{2}\ln 2$

Answer 1

Answer: (D)

Let $I=\underset{\frac{7\pi }{4}}{\overset{\frac{7\pi }{3}}{\int }}\sqrt{{\tan }^{2}x}dx=\underset{\frac{7\pi }{4}}{\overset{\frac{7\pi }{3}}{\int }}|\tan x|dx=\underset{\frac{7\pi }{4}}{\overset{2\pi }{\int }}-(\tan x)dx+\underset{2\pi }{\overset{\frac{7\pi }{3}}{\int }}(\tan x)dx$
$=(\ln (\cos x){)}_{315^{\circ }}^{360^{\circ }}-(\ln (\cos x){)}_{360^{\circ }}^{420^n}=\left(0-\ln \frac{1}{\sqrt{2}}\right)-(\ln \frac{1}{2}-0$
$=\frac{-1}{2}\ln \frac{1}{2}-\ln \frac{1}{2}=\frac{-3}{2}\ln \frac{1}{2}=\frac{3}{2}\ln 2$