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Math Help - Need help with indefinite integral

  1. #1
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    Need help with indefinite integral

    I am pretty sure I need to use integration by parts, but I cannot get anywhere with this.

    \int 2x^3\cos(x^2)\,dx
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  2. #2
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    Quote Originally Posted by blackwrx View Post
    I am pretty sure I need to use integration by parts, but I cannot get anywhere with this.

    \int 2x^3\cos(x^2)\,dx

    Indeed. You can try u=x^2\,,\,u'=2x\,,\,\,v'=2x\cos x\,,\,v=\sin x^2 , so

    \int 2x^3\cos(x^2)\,dx=x^2\sin x^2-\int 2x\sin x^2\,dx = x^2\sin x^2+\cos x^2+C , the last integral being (almost) immediate

    since it is of the form \int f'(x)\sin f(x),dx=-\cos f(x)+C

    Tonio
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  3. #3
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    Hello,

    First substitute u=x^2 and then use once an integration by parts.
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  4. #4
    Super Member Aryth's Avatar
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    You could try a substitution first...

    Let w = x^2, dw = 2x ~dx

    Your integral becomes:

    \int wcos(w) ~dw

    That is easily integrable by parts.

    u = w, du = 1 ~dw, dv = cos(w) ~dw, v = sin(w)

    = wsin(w) - \int sin(w) ~dw

    = wsin(w) + cos(w) + C

    = x^2sin(x^2) + cos(x^2) + C

    And there you go.
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  5. #5
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    Where did the 2xcos(x) come from? What happened to the original 2x^3?
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  6. #6
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    Nevermind, I see with the substitution that dw=2x times the w=x^2 does indeed solve my problem. Thanks guys.
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