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Thread: 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.

    $\displaystyle \int 2x^3\cos(x^2)\,dx$
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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.

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

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

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

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

    Tonio
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  3. #3
    Moo
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    Hello,

    First substitute $\displaystyle u=x^2$ and then use once an integration by parts.
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    You could try a substitution first...

    Let $\displaystyle w = x^2$, $\displaystyle dw = 2x ~dx$

    Your integral becomes:

    $\displaystyle \int wcos(w) ~dw$

    That is easily integrable by parts.

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

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

    $\displaystyle = wsin(w) + cos(w) + C$

    $\displaystyle = x^2sin(x^2) + cos(x^2) + C$

    And there you go.
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  5. #5
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    Where did the $\displaystyle 2xcos(x)$ come from? What happened to the original $\displaystyle 2x^3$?
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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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