First slide

Introduction to integration

Question

$If \int \tan x \tan 2 x \tan 3 x d x = - A \log | \cos 3 x | - B \log | \cos 2 x | + \log | \cos x | + c$

Easy

A

$A = \frac{- 1}{3}, B = \frac{- 1}{2}$
B

$A = \frac{- 1}{3}, B = \frac{1}{2}$
C

$A = \frac{1}{3}, B = \frac{- 1}{2}$
D

None of these

Solution

$\begin{array}{l} Solution. \tan 3 x = \frac{\tan 2 x + \tan x}{1 - \tan 2 x \tan x} \\ ∴ \tan 3 x \tan 2 x \tan x = \tan 3 x - \tan 2 x - \tan x \\ ∴ \int (\tan 3 x - \tan 2 x - \tan x) d x \\ = - \frac{1}{3} \log | \cos 3 x | + \frac{1}{2} \log | \cos 2 x | + \log | \cos x | + c \end{array}$

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