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    \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.
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    (Original post by RDKGames)
    \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.
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    (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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    (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
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    (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'
 
 
 
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