Integrate the following :
$tan x . tan 2 x . tan 3 x = . . .$

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Solution

$t a n (3 x) = t a n (x + 2 x)$
$t a n (3 x) = \frac{t a n x + t a n 2 x}{1 - t a n x . t a n 2 x}$
$t a n (3 x) - t a n x . t a n (2 x) . t a n (3 x) = t a n x + t a n (2 x)$
$∴ t a n x . t a n (2 x) . t a n (3 x) = t a n (3 x) - t a n (x) - t a n (2 x)$ ___________ (1)
$\int t a n x . t a n (2 x) . t a n (3 x) . d x$
$= \int t a n (3 x) - t a n (x) - t a n (x) - t a n (2 x) . d x$ (fomi)
$= \frac{1}{3} l o g$ $| s e c (3 x) | - l o g | s e c x | - \frac{1}{2} l o g | s e c (2 x) | + c$

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