$\int (tan 3 x - tan 2 x - tan x) d x$

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Solution

$\int (t a n (3 x) - t a n (2 x) - t a n x) d x$ $∵ \int t a n x d x = l o g | s e c x | + c$
$= \frac{[l o g (s e c (3 x))]}{3} - \frac{l o g (s e c (2 x))}{2} - \frac{l o g (s e c x)}{1}] + c$
$= \frac{1}{3} l o g (s e c 3 x) - \frac{1}{2} l o g (s e c 2 x) - l o g (s e c x) + c$

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