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Int x^3 sin x

  1. Apr 21, 2007 #1
    I'm trying to integtrate x^3 sin x without using integration by parts. I have set up the equation to either:

    int x^3 cosx dx = (Ax^3 + Bx^2+Cx+D)cos x + K


    int x^3 coxs dx = (Ax^3+Bx)sinx + (Cx^2+D)cosx +K

    but I'm having trouble... any help would be appreciated!
  2. jcsd
  3. Apr 21, 2007 #2


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    Do you want to integrate x3sin(x) or x3cos(x)?
    Why do you think you don't need a term involving sin here?
    Why do you think the polynomial in front of sin doesn't need a square or constant term? Why do you think the polynomial in front of cos doesn't need a cube or linear term?
  4. Apr 21, 2007 #3
    int x^3 cosx dx for this issue

    Thanks for the reply...

    essientially, we need to set this up and are having problems. We have an example of int x^4ex + dx. We get (Ax^4 + Bx^3 + Cx^2 + Ex +F) e^x + k. But we are stuck for finding the starting equation for int x^3 cosx.
  5. Apr 21, 2007 #4

    D H

    Staff: Mentor

    Differentiate and solve for the coefficients.
  6. Apr 21, 2007 #5
    but it's setting up the initial equation that is our road block. We get x^4e^x + dx but not clear on the setting up of X^3 cosx + dx
  7. Apr 21, 2007 #6
    Save yourself the trouble and do integration by parts.
  8. Apr 21, 2007 #7
    we would but we can't use that as per the project instructions X-) Any suggestions?
  9. Apr 21, 2007 #8

    D H

    Staff: Mentor

    How in the world do you get that?

    Do you know how to calculate

    [tex]\frac{d}{dx}\left(\int x^3\,\sin x\, dx\right)[/tex]

    Hint: Its fundamental.
  10. Apr 22, 2007 #9


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    You should note that once you differentiate xncos(x), you'll get:
    (xncos(x))' = nxn - 1cos(x) - xnsin(x), i.e, you have both cos, and sin function on the right, and the highest power of the polynomial is still n.
    If you differentiate xnsin(x), you'll get almost the same.
    (xnsin(x))' = nxn - 1sin(x) + xncos(x)

    Note that, if you differentiate xncos(x), the highest power polynomial will go with sin(x), and vice versa. Note the italic, and underlined part.

    So, if you want to find a function, such that its derivative is x3cos(x) (note that the power of the polynomial term is 3, and is highest), the function should look something like:

    f(x) = (Ax3 + Bx2 + Cx + D) sin(x) + (Ex2 + Fx + G) cos(x)

    Now, if you try to integrate the function xncos(x) by Parts, like this:
    [tex]\int x ^ n \fbox{\cos x} dx = x ^ {n} \fbox{\sin x} - n \int x ^ {n - 1} \sin (x) dx[/tex]
    [tex]= x ^ {n} \fbox{\sin x} + n x ^ {n - 1} \cos(x) - n (n - 1) \int x ^ {n - 2} \cos (x) dx[/tex]
    [tex]= x ^ {n} \fbox{\sin x} + n x ^ {n - 1} \cos(x) - n (n - 1) x ^ {n - 2} \sin (x) + n (n - 1) (n - 2)\int x ^ {n - 3} \sin (x) dx = ...[/tex]
    So, you can see that the result will look something like xn sin(x) + n xn - 1 cos(x) - n (n - 1) xn - 2 sin(x) + ... i.e, one sin, and then one cos, and then back to one sin, and...

    So, you can conclude that the constants B, D, and F in the function f(x) should all be 0, leaving you with:

    f(x) = (Ax3 + Cx) sin(x) + (Ex2 + G) cos(x)

    Can you get it? :)
    Last edited: Apr 22, 2007
  11. Apr 22, 2007 #10


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    Note that it doesn't hurt to have a little overkill. When I saw this problem, I differentiated:

    (Ax4+Bx3+Cx2+Dx+E)sin(x) + (Fx4+Gx3+Hx2+Ix+J)cos(x)

    Even though there are a lot of extra terms, differentiating polynomials and trigonometric functions is easy, and determining these undetermined coefficients ends up being very quick and easy.
  12. Apr 22, 2007 #11
    VietDao29 - THANKS! Your explanation is very clear and was exactly the insight my team needed. We ended up doing just what you suggested before reading your post. The good thing is that we feel confident that we did the project correctly and with your input, it validated our work!

    The interesting part is that we too noticed that the highest exponent went with the opposite trig funciton being integrated. Again, thanks for your reply!
  13. Apr 22, 2007 #12
    AKG - After we tried the problem over and over, when we got what we were expecting, you are exactly right! It was funny, when my team started this, we were apprehensive at best, but now we look at them and laugh 'cause they really are easy when you understand the problem!

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