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Integration help

  1. Oct 9, 2004 #1
    Hi guys... I've been working on this problem for 30 minutes... yes I'm stupid. Seriously I'm thinking right now what the hell I'm going to do. Out of every 4 problems I get 3 wrong and re-do them... I understand where I went wrong but I'm not getting any better. Scary times. :frown:

    Well here it is (btw { = integral of ) :

    Code (Text):
    {  (x)^3 * (cos (x))^3 dx
    the second part is cosine cubed (not x).

    Here's what I got after doing integration by parts 3 times...


    Code (Text):

    x^3 * (sin(x))^2 + x^2 * (cos(x))^2 - 2 * (sin(x))^3
     
     
  2. jcsd
  3. Oct 9, 2004 #2

    arildno

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    1. You ought to rewrite your cubed cosine with trigonometric identities:
    [tex]\cos^{3}(x)=\cos^{2}(x)\cos(x)=\frac{1+\cos(2x)}{2}\cos(x)[/tex]
    [tex]\cos(x)\cos(2x)=\frac{1}{2}(cos(x)+\cos(3x))[/tex]
    That is,
    [tex]\cos^{3}(x)=\frac{1}{4}(3\cos(x)+\cos(3x))[/tex]
    2. Use this expression when doing integration by parts..
     
  4. Oct 9, 2004 #3
    Thanks. :)

    I've never learned those identities.. =\

    btw is there a method to solving it without breaking cosine cubed into smaller pieces?
     
  5. Oct 9, 2004 #4

    arildno

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    Sure; if you are careful with in your calculations, you may do as follows:
    [tex]\cos^{3}(x)=\cos(x)-\sin^{2}(x)\cos(x)[/tex]
    Note that we have the following anti-derivative:
    [tex]\int\sin^{2}(x)\cos(x)dx=\frac{1}{3}\sin^{3}(x)[/tex]
    An analogus decomposition may then be used for [tex]\sin^{3}(x)[/tex]
     
  6. Oct 9, 2004 #5
    awesome. :D

    I have one last question though. ^_^;;

    How to solve the same integral but the cosine is not cubed... that would be:

    { x^3 * cos(x^2) dx

    I also ended up doing integration by parts 3 times on this one.

    *hides under desk*
     
  7. Oct 10, 2004 #6

    arildno

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    [tex]\int{x}^{3}\cos(x^{2})dx=\frac{x^{2}}{2}\sin(x^{2})-\int{x\sin(x^{2})}dx[/tex]
    [tex]\int{x}^{3}\cos(x^{2})dx=\frac{x^{2}}{2}\sin(x^{2})+\frac{\cos(x^{2})}{2}[/tex]
     
  8. Oct 10, 2004 #7
    Thanks for replying. :)

    So you made u = x^3? doesn't that mean du should be 3x^2? :confused:
     
    Last edited: Oct 10, 2004
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