Question

Evaluate: ${\int }_0^{\frac{\pi }{3}}|\tan x-1|dx$

• A. $\frac{\pi }{4}+\frac{1}{2}$
• B. $\frac{\pi }{6}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{4}$

Answer 1

The correct option is B $\frac{\pi }{6}$

${\int }_0^{\frac{\pi }{3}}|\tan x–1|dx$

$={\int }_0^{\frac{\pi }{4}}|\tan x–1|dx+{\int }_{\frac{\pi }{4}}^{\frac{\pi }{3}}|\tan x–1|dx$

$={\int }_0^{\frac{\pi }{4}}(1-\tan x)dx+{\int }_{\frac{\pi }{4}}^{\frac{\pi }{3}}(\tan x–1)dx$

$(\because 0<\tan x<1$ for $0<x<\frac{\pi }{4}$ and $\tan x>1$ for $\frac{\pi }{4}<x<\frac{\pi }{3})$

$=[x+ln|\cos x|{]}_0^{\frac{\pi }{4}}+[-ln|\cos x|-x{]}_{\frac{\pi }{4}}^{\frac{\pi }{3}}$

$(\because \int \tan xdx=-ln|\cos x|)$

$=\frac{\pi }{4}+\log \left(\frac{1}{\sqrt{2}}\right)+(-ln\left(\frac{1}{2}\right)-\frac{\pi }{3}+ln\left(\frac{1}{\sqrt{2}}\right)+\frac{\pi }{4})$

$=\frac{\pi }{6}+ln2+2ln\left(\frac{1}{\sqrt{2}}\right)$

$=\frac{\pi }{6}.(\because 2\ln \left(\frac{1}{\sqrt{2}}\right)=ln\left(\frac{1}{2}\right)=-ln2)$.