Question

${\int }_0^{\pi /3}\left[\sqrt{3}\tan x\right]dx$ equal , where $[.]$ denote greatest integer function

• A. $\frac{\pi }{4}-{\tan }^{-1}\frac{2}{\sqrt{3}}$
• B. $\frac{3\pi }{4}-{\tan }^{-1}\frac{2}{\sqrt{3}}$
• C. $\frac{4\pi }{3}$
• D. $\frac{\pi }{2}-{\tan }^{-1}\frac{2}{\sqrt{3}}$

Answer 1

The correct option is D $\frac{\pi }{2}-{\tan }^{-1}\frac{2}{\sqrt{3}}$

Substitute $\sqrt{3}\tan 0=0,\sqrt{3}\tan \frac{\pi }{3}=3$
$\Rightarrow x=0,1,2,3$
$\therefore {\int }_0^{\frac{\pi }{3}}\left[\sqrt{3}\tan x\right]dx={\int }_0^{\frac{\pi }{6}}\left[\sqrt{3}\tan x\right]dx$
$={\int }_0^{\frac{\pi }{6}}\left[\sqrt{3}\tan x\right]dx+{\int }_{\frac{\pi }{6}}^{{\tan }^{-1}\frac{2}{\sqrt{3}}}\left[\sqrt{3}\tan x\right]dx+{\int }_{{\tan }^{-1}\frac{2}{\sqrt{3}}}^{\frac{\pi }{3}}\left[\sqrt{3}\tan x\right]$
$={\int }_0^{\frac{\pi }{6}}0dx+{\int }_{\frac{\pi }{6}}^{{\tan }^{-1}\frac{2}{\sqrt{3}}}1dx+{\int }_{{\tan }^{-1}\frac{2}{\sqrt{3}}}^{\frac{\pi }{3}}2dx$
$=0+{\tan }^{-1}\frac{2}{\sqrt{3}}-\frac{\pi }{6}+2\left(\frac{\pi }{3}-{\tan }^{-1}\frac{2}{\sqrt{3}}\right)$
$=\frac{\pi }{2}-{\tan }^{-1}\frac{2}{\sqrt{3}}$
Ans: D