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Homework Help: Help with an Integration problem

  1. Dec 4, 2005 #1
    (integration) (sinax)(cosax)dx = (1/2)(integration)[sin2ax)dx
    = (-1/4a)cos2ax = (-1/4a)(1-2(sinax)^2)

    but the correct answer should be (1/2a)sin^ax

    does anybody know what went wrong?
  2. jcsd
  3. Dec 4, 2005 #2


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    Gold Member

    Why should the "correct answer" be (1/2a)sin^ax? How do you know the "correct answer" is actually right?
  4. Dec 4, 2005 #3


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    It's okay

    Sure, you're right, and the back of the book (or Maple or whatever) is also right [assuming you meant (1/2a)sin(ax)^2:

    [tex]\int \sin(ax)\cos(ax)dx = \frac{1}{2}\int \sin(2ax)dx = -\frac{1}{4a} \cos(2ax) +C ,[/tex]

    where C is the constant of integration.

    however, one may instead apply the substitution

    [tex]u=\sin(ax) \Rightarrow du=a\cos(ax)dx[/tex]

    to the given integral like this

    [tex]\int \sin(ax)\cos(ax)dx =\frac{1}{a} \int u du = \frac{1}{2a} u^2 +C = \frac{1}{2a} \sin^{2}(ax) +C[/tex]

    But how could that be? because

    [tex]-\frac{1}{4a} \cos(2ax) =C+\frac{1}{2a} \sin^{2}(ax)[/tex]

    is the half-angle identity from you used [quote: (-1/4a)cos2ax = (-1/4a)(1-2(sinax)^2)] for the proper value of C.
  5. Dec 4, 2005 #4
    thank you very much!!! :)
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