Results 1 to 5 of 5

Math Help - I need help again!

  1. #1
    Junior Member
    Joined
    Jul 2006
    From
    Singapore
    Posts
    55

    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!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    This one can be a little tricky. Use integration by parts.

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

    Therefore:

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

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

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

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

    Can you finish up now?.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jul 2006
    From
    Singapore
    Posts
    55
    Yay thanks so much, galactus, I finally got the answer!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,675
    Thanks
    302
    Awards
    1
    Quote Originally Posted by galactus
    This one can be a little tricky. Use integration by parts.

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

    Therefore:

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

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

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

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

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

    Let y = x^2

    Then:
    \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
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Jul 2006
    From
    Singapore
    Posts
    55
    Thanks for the alternative, Dan. I will try your method when I come across a similar question.
    Follow Math Help Forum on Facebook and Google+


/mathhelpforum @mathhelpforum