$\int \frac{1}{\tan x \tan (60 ° - x) \tan (60 ° + x)} d x =$

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$\frac{1}{3} \log | \cos 3 x | + C$

$3 \log | \sin 3 x | + C$

$\frac{1}{3} \log | \sin 3 x | + C$

$\frac{1}{2} \log | \sin 3 x | + C$

answer is A.

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Detailed Solution

$I = \int \frac{1}{\tan x \tan (60 ° - x) \tan (60 ° + x)} d x$

$= \int \frac{1}{\tan 3 x} d x$ (We know that $\tan x \tan (60 - x) \tan (60 + x) = \tan 3 x$ )

$= \int \frac{\cos 3 x}{\sin 3 x} . d x$

$\sin 3 x = t$

$3 \cos 3 x d x = d t$

$I = \frac{1}{3} \int \frac{1}{t} d t = \frac{1}{3} \log | \sin 3 x | + c$

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