solve ∫tan(x−θ)tan(x+θ)tan 2x dx


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Question Number 54756 by Knight last updated on 10/Feb/19

$s o l v e$ $\int t a n (x - θ) t a n (x + θ) t a n 2 x d x$
	Answered by tanmay.chaudhury50@gmail.com last updated on 10/Feb/19

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