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Math Help - Anti-derivative

  1. #1
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    Anti-derivative

    I am having trouble with this one. e^4x - (cosx)^2

    We just started doing these and im lost lol
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  2. #2
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    Quote Originally Posted by drain View Post
    I am having trouble with this one. e^4x - (cosx)^2

    We just started doing these and im lost lol
    The antiderivative of e^{4x} is (1/4)e^{4x} +C if you use the substitution t=4x.

    The antiderivative of (cos x)^2 can be simplified if you use the half-angle identity,

    (cos x)^2 = (1+ cos (x))/2
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  3. #3
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    Ah thanks for that identity.

    So I came up with

    F = 1/4 (e^(4x)) - 1/2 * x - 1/2 (sinx) + C

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    Quote Originally Posted by drain View Post
    Ah thanks for that identity.

    So I came up with

    F = 1/4 (e^(4x)) - 1/2 * x - 1/2 (sinx) + C

    int{e^4x - (cosx)^2}dx = int{e^4x - (1 + cos2x)/2}dx
    .................................= int{e^4x}dx - (1/2)int{1 + cos2x}dx
    .................................= (1/4)e^4x - (1/2)x - (1/4)sin(2x) + C

    by rite, you should do the integrals of e^4x and cos(2x) by substitution, but they're easy enough to do in your head. apparently you had problems with integrating cos2x. do you think you have it now? i integrated as normal, and then DIVIDED by the derivative of 2x, which is what the substitution method would amount to in this case
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  5. #5
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    Gah... I messed up the identity. Thanks.

    so its (cosx)^2 = (1+cos2x)/2 ? I am bad at trig unfortunately... if you wouldn't mind explaining that identity to me as well that'd be very helpful.

    I found an epxlanation online. But thank you!
    Last edited by drain; April 24th 2007 at 05:02 PM.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by drain View Post
    Gah... I messed up the identity. Thanks.

    so its (cosx)^2 = (1+cos2x)/2 ? I am bad at trig unfortunately... if you wouldn't mind explaining that identity to me as well that'd be very helpful.
    the identity comes directly from the half angle formula, where we plug in 2x for x. do you know the half angle formula?

    anyway, we can get it without the half angle formula

    recall: cos(2x) = (cosx)^2 - (sinx)^2 ............we can change the (sinx)^2
    ....................= (cosx)^2 - (1 - (cosx)^2) = 2(cosx)^2 - 1

    or we can change the (cosx)^2:

    so, cos(2x) = (1 - (sinx)^2) - (sinx)^2 = 1 - 2(sinx)^2


    now, if we take cos(2x) = 2(cosx)^2 - 1 and solve for (cosx)^2 we get:
    cos(2x) = 2(cosx)^2 - 1
    => 1 + cos(2x) = 2(cosx)^2
    => (1 + cos(2x))/2 = (cosx)^2

    if we take cos(2x) = 1 - 2(sinx)^2 and solve for (sinx)^2 we get:
    cos(2x) = 1 - 2(sinx)^2
    => cos(2x) - 1 = -2(sinx)^2
    => 1 - cos(2x) = 2(sinx)^2
    => (1 - cos(2x))/2 = (sinx)^2
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