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Thread: I need help again!

  1. #1
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    I need help again!

    Okay, this question is making me confused.

    (i) State the derivative of sin(x^2).

    I got that correct but I can't get the next part.

    (ii) Find (integrate) x^3 cos (x^2) dx.

    Answer: 1/2 [x^2 sin (x^2) + cos (x^2)] + C.


    Thanks in advance if you could help!
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  2. #2
    Eater of Worlds
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    This one can be a little tricky. Use integration by parts.

    Let $\displaystyle u=\frac{x^{2}}{2},\;\ dv=2xcos(x^{2})dx,\;\ du=xdx,\;\ v=sin(x^{2})$

    Therefore:

    $\displaystyle \int{udv}=uv-\int{vdu}$

    $\displaystyle \int\frac{x^{2}}{2}(2xcos(x^{2}))dx=\frac{x^{2}}{2 }sin(x^{2})-\int{xsin(x^{2})}dx$

    Let $\displaystyle u=x^{2},\;\ \frac{du}{2}=xdx$

    The integral portion becomes $\displaystyle \frac{1}{2}\int{sin(u)}du$

    Can you finish up now?.
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  3. #3
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    Yay thanks so much, galactus, I finally got the answer!
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by galactus
    This one can be a little tricky. Use integration by parts.

    Let $\displaystyle u=\frac{x^{2}}{2},\;\ dv=2xcos(x^{2})dx,\;\ du=xdx,\;\ v=sin(x^{2})$

    Therefore:

    $\displaystyle \int{udv}=uv-\int{vdu}$

    $\displaystyle \int\frac{x^{2}}{2}(2xcos(x^{2}))dx=\frac{x^{2}}{2 }sin(x^{2})-\int{xsin(x^{2})}dx$

    Let $\displaystyle u=x^{2},\;\ \frac{du}{2}=xdx$

    The integral portion becomes $\displaystyle \frac{1}{2}\int{sin(u)}du$

    Can you finish up now?.
    Actually, I find it a bit easier to substitute first:

    Let $\displaystyle y = x^2$

    Then:
    $\displaystyle \int dx \, x^3 cos(x^2) = \frac{1}{2} \int (2xdx) x^2 cos(x^2) = \frac{1}{2} \int dy \, y cos(y)$

    Then you only need integrate by parts once. (I hate integration by parts so I only do as many as I have to.)

    -Dan
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  5. #5
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    Thanks for the alternative, Dan. I will try your method when I come across a similar question.
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