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Homework Help: Integrals: cos(2x)

  1. Jan 14, 2009 #1
    Hello :)

    I'm not sure how to deal with an integral of the form cos(2x), sin(2x), tan(3x), etc. Its the constant that is throwing me off - I know what the antideriv of cosx, sinx or tanx is. My text doesn't seem to explain what to do with the constant and I can't seem to find the info using a direct internet search so I was wondering if someone would please tell me what to do with the constant?

    Help would be greatly appreciated
  2. jcsd
  3. Jan 14, 2009 #2


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    Substitute u=2x, du=2*dx. Do the u integral and substitute back.
  4. Jan 14, 2009 #3

    Thanks, but we actually haven't covered the sub rule yet so I'm wondering if maybe there is a different way to get the answer?

    Updating because I found a similar question in my text...
    The text has a problem showing..

    [tex] \int sin( \Pi x)dx = -(1/ \Pi )cos( \Pi x)dx [/tex]

    So I take it if f(x) = cos(2x)dx then F(x) = (1/2)sin(2x)... but the chpt explaining the sub rule says something about changing the boundaries of a definate integral when using the sub rule.... Would I have to do anything to the boundaries? (I've been asked to solve using FTC where [tex] \int \stackrel{b}{a} = F(b) - F(a) [/tex] ) --I found that problem in the sub rule section which is why I'm confused about the boundaries--
    Last edited: Jan 14, 2009
  5. Jan 14, 2009 #4


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    Ok. The integral of sin(2x) is -(1/2)*cos(2x). You can check that just be differentiating -(1/2)*cos(2x) and seeing that you get sin(2x). The antiderivative -(1/2)*cos(2x) is expressed in terms of the original variable x. So the limits are the same as the original limits. You don't have to change them.
  6. Jan 15, 2009 #5
    Oh right.. that makes sense! Thanks for taking the time to reply :)
  7. Jan 15, 2009 #6


    Staff: Mentor

    I don't think there's a substitution rule, per se. Substitution is nothing more than applying the chain rule of differentiation to antidifferentiation.
    To expand on what I said above, if g(x) = sin(2x), then g'(x) = 2cos(2x). Looking at things from the opposite direction,
    [tex]\int 2cos(2x) dx = sin(2x) + C[/tex]
    [tex]\int cos(2x) dx = 1/2 sin(2x) + C_1[/tex]
    where C1 = C/2.

    Here's a slightly more complicated problem:
    [tex]\int xcos(x^2) dx[/tex]

    For this problem, I can see that part of the integrand is cos(x^2), so I think of this as being cos(u(x)), where u(x) = x^2. I see also that du/dx = 2x, so it's helpful that the other factor in the integrand is x.

    So x*cos(x^2)*dx = 1/2 * du/dx * cos(u) * dx = 1/2 * cos(u)*du

    I know that
    [tex]\int cos(u) du = sin(u) + C_1[/tex]

    [tex]\int 1/2*cos(u) du = 1/2*sin(u) + C[/tex]

    = 1/2*sin(x^2) + C
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