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Math Help - Integration

  1. #1
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    Integration

    \int tan(x)tan(2x)tan(3x)dx
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  2. #2
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    Re: Integration

    Answer: (1/2)[ln(2(cosx)^2-1]+(2/3)ln(cosx)-(1/3)[4(sinx)^2-1] + Constant
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  3. #3
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    Re: Integration

    Actually, this is a prove question, I really need steps

    thanks
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  4. #4
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    Re: Integration

    I need to leave ....
    But try tan(2x)=2tanx/(1-(tanx)^2) , tan(3x)=(3tanx-(tanx)^3)/(1-3(tanx)^2)
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  5. #5
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    Re: Integration

    Smoke..
    Just mow I came back home and I started to solve your integral...
    I have discovered a very interesting result
    believe it or not (tanx)(tan2x)(tan3x) = tan(3x)-tan(2x)-tanx
    indeed :
    tan(3x) = (tanx+tan(2x))/(1-tanxtan(2x))
    this implies : tan(3x)(1-tanxtan(2x))=tanx+tan(2x) and after all simplifications we obtain : tan(3x)-tan(2x)-tan(x) = tanxtan(2x)tan(3x)

    therefore your integral now is simple to calculate:
    since integral(tanx)dx = -ln(cosx)
    integral(tan(2x))dx = (-1/2)ln(cos(2x))
    and Integral (tan(3x))dx = (-1/3)ln(cos(3x))
    you can verify all these and check the calculations..

    MINOAS
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  6. #6
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    Re: Integration

    Thank you!
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  7. #7
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    Re: Integration

    ok
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