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How would I approach this integral?

∫0π61−2tan(2θ)tan(4θ) dθ\displaystyle\int^\frac{\pi}{6}_0 1-\frac{2tan(2\theta)}{tan(4\theta)}\ d\theta

The only idea I'm having is to use double angle but knowing it for tan I can also see it going insane and being too long-winded.
Original post by RDKGames
∫0π61−2tan(2θ)tan(4θ) dθ\displaystyle\int^\frac{\pi}{6}_0 1-\frac{2tan(2\theta)}{tan(4\theta)}\ d\theta

The only idea I'm having is to use double angle but knowing it for tan I can also see it going insane and being too long-winded.


Unless I'm missing something using the double angle formula makes it nice and neat.
Original post by TeeEm
Very quick and not long winded


Original post by 133221333123111
Unless I'm missing something using the double angle formula makes it nice and neat.


I wouldn't mind it for cos or sin but with tan it really doesn't look that way. Anyhow, I'll attempt it and see how it goes.
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math42
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Original post by RDKGames
I wouldn't mind it for cos or sin but with tan it really doesn't look that way. Anyhow, I'll attempt it and see how it goes.


in particular tan(4theta) = tan(2theta + 2theta)
works out nice at a glance
Original post by 133221333123111
Unless I'm missing something using the double angle formula makes it nice and neat.


Original post by TeeEm
Very quick and not long winded


It does indeed turn out to be quite more simple that I first thought, well thank you for the confidence, I have a very bad case of 'getting-scared-when-seeing-trig-in-integrals-with-multiples-angles'
(edited 7 years ago)

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