Question

The value of ${\int }_0^{\frac{\pi }{3}}log\left(1+\sqrt{3}tanx\right)dx$ is equal to

• A. $\pi log2$
• B. $\frac{\pi }{2}log2$
• C. $\frac{\pi }{3}log2$
• D. $\frac{\pi }{4}log2$

Answer 1

Answer: (C)

Let, $I={\int }_0^{\frac{\pi }{3}}\log (1+\sqrt{3}\tan x)dx$
$\begin{array}{l}I={\int }_0^{\frac{\pi }{3}}\log \left[1+\sqrt{3}\tan \left(\frac{\pi }{3}-x\right)\right]dx\\ ={\int }_0^{\frac{\pi }{3}}[\log \left(1+\sqrt{3}\tan \left(\frac{\sqrt{3}-\tan x}{1+\sqrt{3}\tan x}\right)\right]dx\\ ={\int }_0^{\frac{\pi }{3}}\log \left(\frac{1+\sqrt{3}\tan x+3-\sqrt{3}\tan x}{1+\sqrt{3}\tan x}\right)dx\\ \Rightarrow I={\int }_0^{\frac{\pi }{3}}(\log 4-\log (1+\sqrt{3}\tan x))dx\\ \Rightarrow I=\log 4\cdot \frac{\pi }{3}-I\\ \Rightarrow I=\frac{\pi }{6}\log 4=\frac{\pi }{3}\log 2.\end{array}$