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Math Help - Integrating by parts + substitution

  1. #1
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    Integrating by parts + substitution

    Hey, could I get some guidance on how to do this question:

    Integrate cos(x)sin(x)e^(cos(x))

    I figure you'd also have to use some trig identities here, correct?

    My first approach was:

    let u = e^(cos(x))
    so du/dx = -sin(x)e^cos(x)

    let dv/dx = sin(x)cos(x) = 0.5sin(2x)
    so v = -0.25cos(2x)

    => uv - integral of -0.25cos(2x)e^cos(x)
    .. and yay, we're back to square one practically
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  2. #2
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    Re: Integrating by parts + substitution

    Hey 99.95.

    Try the substitution u = cos(x) instead and get the integral in terms of u and then note that you can integrate by parts.
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  3. #3
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    Re: Integrating by parts + substitution

    Integrate = I =cos(x)sin(x)e^(cos(x))

    let u = cos(x)
    du/dx = -sin(x)
    du = sin(x)dx

    => -I = integral of (ue^u)du

    let f = u let dg/du = e^u
    df/du = 1 therefore g = e^(u)/u

    again:
    ug - integral of (e^(u)/u) ????

    which gives us e^u - e^(u)/u^2 (edit: I realise this is wrong)

    = e^cos(x) - e^cos(x)/cos^2(x) + c

    Still wrong, i believe, but i got further!
    does this forums support latex?

    where was my mistake?
    Last edited by 99.95; November 4th 2012 at 01:15 AM.
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  4. #4
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    Re: Integrating by parts + substitution

    Quote Originally Posted by 99.95 View Post
    Integrate = I =cos(x)sin(x)e^(cos(x))

    let u = cos(x)
    du/dx = -sin(x)
    du = sin(x)dx

    => -I = integral of (ue^u)du

    let f = u let dg/du = e^u
    df/du = 1 therefore g = e^(u)/u

    again:
    ug - integral of (e^(u)/u) ????

    which gives us e^u - e^(u)/u^2 (edit: I realise this is wrong)

    = e^cos(x) - e^cos(x)/cos^2(x) + c

    Still wrong, i believe, but i got further!
    does this forums support latex?

    where was my mistake?
    Last time I checked, the integral of e^u is e^u, not (e^u)/u...
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  5. #5
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    Re: Integrating by parts + substitution

    If dg/du = e^u then g = e^u, not [e^(u)]/u.

    EDIT: yeah, like Prove It said.
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