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Math problem

  1. Nov 7, 2003 #1
    Ok here it is,

    Show that the integral of


    = -(1/2)ln(2sin(2θ)+2)+C

    Now what I did was let u=sin(θ)+cos(θ) so that du=cos(θ)-sin(θ)dθ or that du=-(sin(θ)-cos(θ))dθ

    Now I got the integral of


    And then both of the sin(θ)-cos(θ) cancels out and I was left with the integral of


    which is -ln|u|+C , plugging back what u was into that I got

    which I cannot figure out to be equal to what he said it should be.

    The only thing I can think of is the C's being different and that causing them to be equal. Any ideas?
  2. jcsd
  3. Nov 7, 2003 #2


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    Notice that their solution has a (1/2) out front. Try multiplying your solution by 2/2. (since 2/2 = 1, this doesn't change the solution)

    Now, leave the (1/2) outside the logarithm, but use the power rule (n log a = log (a^n)) to pull the 2 inside the logarithm, and see if you can manipulate the result to get their answer. You'll probably be off by a multiplicative constant in the logarithm, but then you can use the fact that log ab = log a + log b to pull it out and absorb it into C.
  4. Nov 7, 2003 #3
    Of course mutliply by one. I always forget to multiply by 1, but I never forget to add 0. Thanks


    Ok I now got to this answer


    Now can I add -(1/2)ln(2)? And then pull the 2 inside the original log and make a new constant C? to get the answer I wanted of -(1/2)ln(2sin(2θ)+2)+C?
    Last edited: Nov 7, 2003
  5. Nov 7, 2003 #4


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