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Maths - integration

  1. May 14, 2004 #1
    \int\frac{\sin^2 x}{e^x}
    = \int\frac{1}{2e^x} - \frac{\cos2x}{2e^x} dx
    = - \frac{1}{2}e^{-x} - \frac{1}{2}\int\frac{\cos 2x}{e^x} dx
    im trying to use the integration by parts method, but seems like im stuck there, doesnt reach the answer :-

    [tex] -\frac{e^{-x}}{2} + \frac{e^{-x}}{10} ( cos 2x - 2 sin 2x ) + c [/tex]
    Last edited: May 14, 2004
  2. jcsd
  3. May 14, 2004 #2

    Dr Transport

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    write [tex] \cos(2x) = \frac{e^{2ix}+e^{-2ix}}{2} [/tex] then integrate......is should work out.
  4. May 14, 2004 #3
    i never come across that formula. can u explain?
  5. May 14, 2004 #4
    ok..... i already know how to prove it. but may i know how u come across that formula?
  6. May 14, 2004 #5

    Dr Transport

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    it is the expansion of the trigonometric functions in terms of complex exponentials......
  7. May 14, 2004 #6
    It's known as the Euler Formula.

    [tex]e^{i\theta} = \cos{\theta} + i\sin{\theta}[/tex]

    And, using the even/oddness of cos/sin,

    [tex]e^{-i\theta} = \cos{\theta} - i\sin{\theta}[/tex]

    Which can then be combined to solve for either the sine or cosine.

    Last edited: May 14, 2004
  8. May 14, 2004 #7
    I wouldn't expect that he'd be required to know anything about complex numbers for this...
  9. May 15, 2004 #8
    yeah, it doesnt. anyways thanks for the suggestion.
    Last edited: May 15, 2004
  10. May 15, 2004 #9
    bump bmp bump
    Last edited: May 15, 2004
  11. May 17, 2004 #10

    Just outta interest, how do u integrate that? do you just integrate as per real numbers or are their any special complex techniques?
  12. May 17, 2004 #11

    Dr Transport

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    just integrate with respect to [tex] x[/tex], the complex number [tex] i [/tex] is a constant.....
  13. May 17, 2004 #12
    Repeated application of integration-by-parts is the suggested route. Denian set up the problem correctly, now just needs to proceed carefully:

    Let the last integral represent [itex]I_1[/itex], ie,

    [tex]I_1 = \frac{1}{2}\int\frac{\cos 2x}{e^x} dx[/tex]

    and select

    [tex]u = \cos 2x [/tex]

    [tex]dv = e^{-x} dx[/tex]


    [tex]2I_1 = \int u dv = uv - \int v du [/tex]

    [tex]2I_1 = -e^{-x} \cos 2x - 2 \int e^{-x} sin 2x dx [/tex]

    Now repeat this process with the last integral in expression for I1, ie, let

    [tex]I_2 = \int e^{-x} sin 2x dx[/tex]

    Calculate via integration-by-parts and the resulting expression for [itex]I_2[/itex] should yield a trailing integral which you'll reconize as a multiple of [itex]I_1[/itex]. Substitute "[itex]I_1[/itex]" for this integral and collect terms to solve for "[itex]I_1[/itex]". Substitute this into the original expression and your done.
  14. May 17, 2004 #13


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    [tex]\int\frac{\sin(x)^2}{e^x} dx =[/tex]
    [tex]\int e^{-x} \sin(x)^2 dx[/tex]
    [tex]-\int e^u \sin(u)^2 du = [/tex]
    [tex]-(e^u \sin(u)^2 - \int e^{u} 2 \sin(u) \cos(u) du)=[/tex]
    [tex]-(e^u \sin(u)^2 - \int e^{u} 2 \sin(2u) du)=[/tex]
    [tex]-(e^u \sin(u)^2 + e^{u} \cos(2u) + \int e^{u} \cos(2u))=[/tex]
    [tex]-(e^u \sin(u)^2 + e^{u}(\cos^2(u) - \sin^2(u)) + \int e^{u} \cos(2u) du)=[/tex]
    [tex]-e^u \cos^2(u) - \int e^{u} \cos^2(u)du + \int e^{u} \sin^2(u) du =[/tex]
    [tex]-e^u \cos^2(u) - \int e^{u} du + 2 \int e^{u} \sin^2(u) du=[/tex]
    so we have
    [tex]\int e^{u} \sin^2{u} du = e^u \cos^2(u) + \int e^{u} du - 2 \int e^{u} \sin^2{u} du[/tex]
    [tex]\int e^{u} \sin^2{u} du = \frac{e^u \cos^2(u) + e^u)}{3}[/tex]
    [tex]\int\frac{\sin(x)^2}{e^x} dx = -\int e^u \sin(u)^2 du = -\frac{e^u \cos^2(u) + e^u}{3} + C[/tex]
    [tex]\int\frac{\sin(x)^2}{e^x} dx = - \frac{\cos^2(x)+1}{3e^x} + C[/tex]

    Of course, you should check my work etc.
  15. May 18, 2004 #14
    Why does my brain just decide to switch of sometimes...

    Stupid, stupid, stupid! Of course i is a constant!
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