Question

If $I=\int \frac{\tan x}{\tan 3x}dx$, then $I$ equals

• A. $x-\frac{1}{\sqrt{3}}\log \left|\frac{\sqrt{3}+\tan x}{\sqrt{3}-\tan x}\right|+C$
• B. $x+\frac{1}{\sqrt{3}}\log \left|\frac{\sqrt{3}+\tan x}{\sqrt{3}-\tan x}\right|+C$
• C. $x-\frac{1}{\sqrt{3}}\log \left|\frac{3-\tan x}{3+\tan x}\right|+C$
• D. $x+\frac{2}{\sqrt{3}}\log \left|\frac{3-\tan x}{3+\tan x}\right|+C$

Answer 1

The correct option is A $x-\frac{1}{\sqrt{3}}\log \left|\frac{\sqrt{3}+\tan x}{\sqrt{3}-\tan x}\right|+C$

$I=\int \frac{\tan x}{\tan 3x}dx=\int \frac{\tan x(1-3{\tan }^{2}x)}{\tan x(3-{\tan }^{2}x)}dx$
Substitute $\tan x=t$
$I=\int \frac{1-3t^2}{(3-t^2)(1+t^2)}dt=\int (\frac{1}{t^2-1}-\frac{2}{(3-t^2)})dt$
$=-{\tan }^{-1}t-\frac{1}{\sqrt{3}}\log \left|\frac{\sqrt{3}+t}{\sqrt{3}-t}\right|+c$