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    Okay, well actually there is only one but HEAR ME OUT plz.

    So I was trying to solve

    I did the idea of integrating a first order differential to get me back to the original function (+C!!!!!)

    So I did y=-1/4 * csc^2(2x)

    which when I put into WA, is identical to the integrand - But it's not the answer.

    Why?!
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    (Original post by DrSebWilkes)
    Okay, well actually there is only one but HEAR ME OUT plz.

    So I was trying to solve

    I did the idea of integrating a first order differential to get me back to the original function (+C!!!!!)

    So I did y=-1/4 * csc^2(2x)

    which when I put into WA, is identical to the integrand - But it's not the answer.

    Why?!
    I haven't checked this, but I'll assume you have everything right so far. When you say this isn't the answer, do you mean it is not the given answer? As in the mark schme or text book answer? Because if the differential of your answer is the original integrand, then it is correct. I expect that you will need to do some trig identity work to verify that your answer and their answer are, in fact, the same.
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    (Original post by DrSebWilkes)
    Okay, well actually there is only one but HEAR ME OUT plz.

    So I was trying to solve

    I did the idea of integrating a first order differential to get me back to the original function (+C!!!!!)

    So I did y=-1/4 * csc^2(2x)

    which when I put into WA, is identical to the integrand - But it's not the answer.

    Why?!
    Note cosec^2 and cot^2 only differ by a constant, and you get an arbitrary constant when you integrate...
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    (Original post by Pangol)
    I haven't checked this, but I'll assume you have everything right so far. When you say this isn't the answer, do you mean it is not the given answer? As in the mark schme or text book answer? Because if the differential of your answer is the original integrand, then it is correct. I expect that you will need to do some trig identity work to verify that your answer and their answer are, in fact, the same.
    Yeah as I said I went on WA and differentiated my "answer" and got exactly the integrand so I thought "brilliant" and left it.

    I check the back and the answer is -1/4 cot^2(2x)

    And WA says they are not equal (IE: not indentical) EDIT: ANNNND WA says the integral of the integrand is the back-of-book answer. Mystery deepens

    Soooo ... yeah :| annoying this is I don't know many identities atm but I am trying to learn them. I'm trying to cram in C4 before November 2nd, you see
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    (Original post by DFranklin)
    Note cosec^2 and cot^2 only differ by a constant, and you get an arbitrary constant when you integrate...
    so, with a +C, am I right?
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    I presume where you mention "WA" you mean Wolfram Alpha.

    Wolfram Alpha gives the antiderivative as  -(1/4) \cot ^{2} (2x) + C, (where C is a constant).

    As DFranklin says,  \cot^{2} (x) and  \mathrm{cosec} ^{2} (x) only differ by a constant.

    Note, as under "Alternate forms of the integral", it lists:

     - \dfrac{ \cos^{2} (2x) }{ 4 \sin^{2} (2x) } + C, then we can use the  \sin^{2} (x) and  \cos^{2} (x) identity.
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    (Original post by simon0)
    I presume where you mention "WA" you mean Wolfram Alpha.

    Wolfram Alpha gives the antiderivative as  -(1/4) \cot ^{2} (2x) + C, (where C is a constant).

    As DFranklin says, cot^{2} (x) and cosec^{2} (x) only differ by a constant.

    Note, as under "Alternate forms of the integral", it lists:

     - \dfrac{ \cos^{2} (2x) }{ 4 \sin^{2} (2x) } + C.
    As I mentioned, yeah.

    But the point still remains that if you differentitate csc^2 (2x) * -1/4 you get the integrand.

    So maybe that +C at the end takes care of it?
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    (Original post by DrSebWilkes)
    As I mentioned, yeah.

    But the point still remains that if you differentitate csc^2 (2x) * -1/4 you get the integrand.

    So maybe that +C at the end takes care of it?
    Yes.
 
 
 
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