Question

find the value of the integral of the greatest integer function of root 3 into tanx from 0 to π/3

Answer 1

So for interval of $[0, \frac{\pi}{3}]$
tan x lies in $[0,\sqrt{3} ]$
now $\sqrt{3} tan(x)$ lies in $[0,3]$
I denote greatest Integer function as y(x)
So y( $\sqrt{3} tan(x)$) will have values as 0,1,2,3 for particular corresponding intervals
[0,0],
(0,$tan^{-1}( \frac{1}{ \sqrt{3} } )$)
($tan^{-1}( \frac{1}{ \sqrt{3} } )$,$tan^{-1}( \frac{2}{ \sqrt{3} } )$)
($tan^{-1}( \frac{2}{ \sqrt{3} } )$,$\frac{\pi}{3}$)

So now compute summation of integrals over all intervals
[tex] \int\limits^ \frac{\pi}{3} _0 {y( \sqrt{3} tan(x) )} \, dx = \\ \int\limits^ {tan^{-1}( \frac{1}{ \sqrt{3} } )} _0 {1} \, dx + \\ \int\limits^ {tan^{-1}( \frac{2}{ \sqrt{3} } )} _{tan^{-1}( \frac{1}{ \sqrt{3} } )} {2} \, dx + \\ \int\limits^ \frac{\pi}{3} _{tan^{-1}( \frac{2}{ \sqrt{3} } )} {3} \, dx = \\ tan^{-1}( \frac{1}{ \sqrt{3} } ) + 2tan^{-1}( \frac{2}{ \sqrt{3} } ) - 2tan^{-1}( \frac{1}{ \sqrt{3} } ) \\ \pi - 3tan^{-1}( \frac{2}{ \sqrt{3} } ) [/tex]

$= \pi - tan^{-1}( \frac{2}{ \sqrt{3} } ) - tan^{-1}( \frac{1}{ \sqrt{3} } ) = \frac{5 \pi}{6} - tan^{-1}( \frac{2}{ \sqrt{3} } )$

Answer is B)