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Integrate : $\int \tan x \tan 2x \tan 3x dx$

$(a)\;\frac{1}{3}\log | \sec (3x)|-\frac{1}{2} \log | \sec 2x| - \log | \sec x | +c \\(b)\;\frac{4}{3}\log | \sec (3x)|-\frac{1}{2} \log | \sec 2x| - \log | \sec x | +c \\(c)\;\frac{1}{3}\log | \sec (3x)|+\frac{1}{2} \log | \sec 2x| - \log | \sec x | +c \\ (d)\;\frac{7}{3}\log | \sec (3x)|-\frac{1}{2} \log | \sec 2x| - \log | \sec x | +c$

1 Answer

$\int \tan x \tan 2x \tan 3x dx$-----(i)
$\tan 3x= \tan (x+2x)$
$\qquad= \large\frac{\tan x +\tan 2x}{1- \tan x \tan 2x}$
$\bigg\{ \tan (A+B)= \large\frac{\tan A+ \tan B}{1- \tan A \tan B} \bigg\}$
$\tan 3x -\tan x -\tan 2x =\tan x \tan 2x \tan 3x$-------(ii)
From equation (i) and (ii)
$\int \tan 3x- \tan x- \tan 2x dx$
=>$\large\frac{1}{3}$$\log | \sec (3x)|-\frac{1}{2} \log | \sec 2x| - \log | \sec x | +c$
Hence a is the correct answer.
answered Dec 24, 2013 by meena.p