First slide

Introduction to integration

Question

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

Easy

A

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

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

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

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

Solution

$\int \frac{1}{\tan 3 x} d x = \int \cot 3 x d x$

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

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