ϕ = ∫ tan (x+(π/3))tan 3x tan (2x−(π/3)) dx =?


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Question Number 149112 by bramlexs22 last updated on 03/Aug/21

$φ = \int \tan (x + \frac{π}{3}) \tan 3 x \tan (2 x - \frac{π}{3}) dx = ?$
Commented by som(math1967) last updated on 03/Aug/21

$t a n 3 x = t a n (x + \frac{π}{3} + 2 x - \frac{π}{3})$ $t a n 3 x = \frac{t a n (x + \frac{π}{3}) + t a n (2 x - \frac{π}{3})}{1 - t a n (x + \frac{π}{3}) t a n (2 x + \frac{2 π}{3})}$ $∴ t a n 3 x t a n (x + \frac{π}{3}) t a n (2 x - \frac{π}{3})$ $= t a n 3 x - t a n (x + \frac{π}{3}) - t a n (2 x - \frac{π}{3})$
Commented by bramlexs22 last updated on 03/Aug/21

$yes thanks$
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