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Math Help - Very basic antidifferentiation problem

  1. #1
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    Very basic antidifferentiation problem

    I still don't know the math symbols, so please bear with me!

    I have the big S thing over sin(2X)*DX, its antidifferentiation is -(cos(2X))/2 + C

    I have completely lost where that over 2 comes from. Wolfram alpha is not helpful. I realize that somewhere we pull a 1/2 out in front of everything, but I don't know where.

    Thanks, I hope I'm making SOME sense.
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  2. #2
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    Remember that

     \int \sin nx ~dx = -\frac{1}{n}\cos nx +C

    Therefore

     \int \sin 2x ~dx = -\frac{1}{2}\cos 2x +C
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  3. #3
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    Quote Originally Posted by Wolvenmoon View Post
    I still don't know the math symbols, so please bear with me!

    I have the big S thing over sin(2X)*DX, its antidifferentiation is -(cos(2X))/2 + C

    I have completely lost where that over 2 comes from. Wolfram alpha is not helpful. I realize that somewhere we pull a 1/2 out in front of everything, but I don't know where.

    Thanks, I hope I'm making SOME sense.
    formally ...

    \int \sin(2x) \, dx<br />

    u = 2x

    du = 2 \, dx

    \frac{1}{2} \int \sin(2x) \cdot 2 \, dx

    substitute ...

    \frac{1}{2} \int \sin(u) \, du<br />

    integrate ...

    -\frac{1}{2} \cos(u) + C<br />

    back-substitute ...

    -\frac{1}{2} \cos(2x) + C


    informally ...

    \int \sin(kx) \, dx

    the antiderivative of \sin(kx) has to be a form of -\cos(kx), but taking the derivative of -\cos(kx) yields \sin(kx) \cdot k because of the chain rule ... now, what do you think will get rid of the constant factor k ?
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  4. #4
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    Okay Skeeter, I went back over this and there are two steps in particular that have me. When the 1/2 shows up in front of everything, how did it get there and why?

    My instructor has taken a completely different approach to this than my book did, hers is an easier way but I think I've managed to drop something somewhere.
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  5. #5
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    Quote Originally Posted by Wolvenmoon View Post
    When the 1/2 shows up in front of everything, how did it get there and why?
    Find \frac{d}{dx}\left(-\frac{1}{n}\cos nx +C\right)

    That should make things clearer.
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  6. #6
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    When you are substituting into \int sin(2x) dx, you have to substitute for every "x", even the one is "dx".

    As skeeter said, he let u= 2x. Differentiating both sides of that, du= 2 dx.

    There are two ways of thinking about that but they give the same result:

    1) Multiply and divide the integral by 2-
    \int sin(2x)\left(\frac{1}{2}\right)(2x) dx= \frac{1}{2}\int sin(2x) 2dx so your integral becomes \frac{1}{2}\int sin(u)du.

    2) Divide both sides of du= 2 dx by 2 to get \frac{1}{2} du= dx and replace dx by that-
    \int sin(u) \frac{1}{2} du= \frac{1}{2}\int sin(u) du.

    Notice that it is the fact that " \frac{1}{2}" is a constant that allows us to move it in or out of the integral at will.
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  7. #7
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    So essentially that's 2/2 or a form of 1 that we're adding into the picture to force du = 2dx, breaking up into 2/1 and 1/2, and moving the 1/2 over while we work?


    Okay. Think I've cleared the brick wall here. Thanks!
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