Mathematics - Definite Integration Question with Solution | TestHub

${\int }_0^{\frac{\pi }{3}}[\sqrt{3}\tan x]dx={\int }_0^{\frac{\pi }{6}}[\sqrt{3}\tan x]dx+{\int }_{\frac{\pi }{6}}^{{\tan }^{-1}\frac{2}{\sqrt{3}}}[\sqrt{3}\tan x]dx+{\int }_{{\tan }^{-1}\frac{2}{\sqrt{3}}}^{\frac{\pi }{3}}[\sqrt{3}\tan x]dx$

$$\begin{array}{rlr}& =\frac{\pi }{2}-{\tan }^{-1}\frac{2}{\sqrt{3}}={\sin }^{-1}(\sqrt{3/7})& \Rightarrow k=3/7\end{array}$$

$$\begin{array}{rlr}& \therefore {\int }_0^{2k}[\tan x]dx={\int }_0^{\frac{6}{7}}[\tan x]dx={\int }_0^{\frac{\pi }{4}}[\tan x]dx+{\int }_{\frac{\pi }{4}}^{\frac{6}{7}}[\tan x]dx& =\frac{6}{7}-\frac{\pi }{4}\end{array}$$