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

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

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

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

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

answer is A.

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Detailed Solution

$\int \frac{1}{\tan x \tan (60 ° - x) \tan (60 ° + x)} d x = \int \frac{1}{\tan 3 x} d x (\tan θ . \tan (60 - θ) \tan (60 + θ) = \tan 3 θ) = \int c o t 3 x . d x = \frac{\log |\sin 3 x|}{3} + C$

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