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Int(exp(-t) * (cos(t))^2)

  1. Oct 5, 2008 #1
    [tex] \int^{t}_{0}e^{-s}cos^2(s)ds [/tex]

    let [tex] u = e^{-s}, \ du = -e^{-s}, \ dv = cos^{2}(s)ds, \ v = \frac{1}{2}s + \frac{1}{4}sin(2s) [/tex]


    [tex] \int^{t}_{0}e^{-s}cos^2(s)ds =\left[e^{-s}(\frac{1}{2}s - \frac{1}{4}sin(2s)\right]^{t}_{0} + \int^{t}_{0}e^{-s}(\frac{1}{2}s-\frac{1}{4}sin(2s)ds [/tex]

    let [tex] u = e^{-s}, \ du = -e^{-s}ds, \ dv = \frac{1}{2}s-\frac{1}{4}sin(2s)ds, \ v = \frac{1}{4}s^{2} + \frac{1}{8}cos(2s) [/tex]


    [tex] = e^{-t}\left(\frac{1}{2}t - \frac{1}{4}sin(2t)\right) + \left[e^{-s}\left(\frac{1}{4}s^{2} + \frac{1}{8}cos(2s)\right)\right]^{t}_{0} + \int^{t}_{0}e^{-s}\left(\frac{1}{4}s^{2} + \frac{1}{8}cos(2s)\right) [/tex]

    How to proceed? I can't seem to get this

    [tex]\int^{t}_{0}e^{-s}\left(\frac{1}{4}s^{2} + \frac{1}{8}cos(2s)\right) [/tex]

    to equal this

    [tex] \int^{t}_{0}e^{-s}cos^2(s)ds [/tex]

    So I'm not sure what to do next.

    Any suggestions?
  2. jcsd
  3. Oct 5, 2008 #2


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    Science Advisor

    Re: int(exp(-t)*(cos(t))^2)

    I would be inclined to do it the other way around: let u= cos2(s), dv= e-s.

    That, together with the fact that sin(2s)= 2 sin(s)cos(s), should make the second integration by parts easy.
  4. Oct 5, 2008 #3
    Re: int(exp(-t)*(cos(t))^2)

    Ok thanks, I'll give that a try.
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