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Math Help - [SOLVED] What am I doing wrong?

  1. #1
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    [SOLVED] What am I doing wrong?

    This has been bugging me for a long time now... Can anyone please help me?
    Integrate dx/(sin(x) + cos(x))




    I used the identy that t= tan(x/2)
    So following the sin and cos substitutaion I have
    dx = 2dt/1+t^2
    sin(x) = 2t/(1+t^2)
    cos(x) = (1-t^2)/(1+t^2)
    thus we have
    Integral (2dt/(1+t^2))/((2t + 1-t^2)/(1+t^2)
    this simplifies to
    Integral 2dt/(2t+1-t^2)
    then factor down into
    Integral -2dt/(t-1)^2 -2
    t=sqrt(2)sec(a) +1
    Integral -2dt/2(sec(a)^2 -1)
    Integral -dt/(tan^2(a))

    Am I way off?? there seems to be way to many substitutions or something... I would appreciate it very much!
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  2. #2
    MHF Contributor matheagle's Avatar
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    I try to avoid the half angle formula.
    Instead I would multiply by the conjugate on top and bottom.
    It works, but you need to break up the fraction into two pieces
    and then there's a little partial fraction problem to solve, but it should work.
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  3. #3
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    Wish i would have noticed that!! Thanks! I will try it that way now.
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  4. #4
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    working the other way, this now becomes the answer

    Integral (sinx - cosx)dx / (sin^2(x) - cos^2(x))
    Integral (sinx - cosx)dx/(2sin^2(x) - 1)
    u = sinx
    du = cosxdx
    Integral (-udu)/(2u^2 - 1)
    -(1/4)ln(1-2u^2)
    -(1/4)ln(1-2sinx^2)
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  5. #5
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by codex123 View Post
    Wish i would have noticed that!! Thanks! I will try it that way now.
    I reviewed Salas/Hille a couple of years ago and they too left out that trick. I mentioned it, but they never used my suggestion. What's weird was that 30 years earlier I used that book as an undergraduate and now they pay me to review the same book.
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  6. #6
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by codex123 View Post
    working the other way, this now becomes the answer

    Integral (sinx - cosx)dx / (sin^2(x) - cos^2(x))
    Integral (sinx - cosx)dx/(2sin^2(x) - 1)
    u = sinx
    du = cosxdx
    Integral (-udu)/(2u^2 - 1)
    -(1/4)ln(1-2u^2)
    -(1/4)ln(1-2sinx^2)

    Be careful, your numerator in (-udu)/(2u^2 - 1)
    Looks wrong.
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