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Math Help - Trying to integrate cos 5x

  1. #1
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    Trying to integrate cos 5x

    can some please explain this to me:

    if dv = cos 5x dx, how does v = 1/5 sin 5x ??

    --i get something totally different (5/2 x^2 sin 5x )

    Thanks for any help!
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by adam21 View Post
    can some please explain this to me:

    if dv = cos 5x dx, how does v = 1/5 sin 5x ??

    --i get something totally different (5/2 x^2 sin 5x )

    Thanks for any help!
    Here
    Attached Thumbnails Attached Thumbnails Trying to integrate cos 5x-int.gif  
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    You can check your answers by differentiating the answer you got when you integrate.

    If you differentiate 5/2 x^2 sin 5x you don't get cos(5x), but if you differentiate (1/5)sin(5x) you do.

    i'm not sure where you made your mistake, but i think you were mixing up the product rule somewhere in all this, but you don't have a product
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by adam21 View Post
    can some please explain this to me:

    if dv = cos 5x dx, how does v = 1/5 sin 5x ??

    --i get something totally different (5/2 x^2 sin 5x )

    Thanks for any help!
    Note that the 5x is the argument of the cosine function; it is not mulitpling anything.

    Int[cos(5x)dx]

    Let y = 5x --> dy = 5dx

    Thus:
    Int[cos(5x) dx] = Int[cos(y) * (1/5) dy] = (1/5)*Int[cos(y) dy]

    = (1/5)*sin(y) + C = (1/5)*sin(5x) + C

    -Dan
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