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X * sin x * sin2x integral

  1. Oct 24, 2003 #1
    As a minor step in a quantum mechanics problem I need to integrate x * sin(pi*x/a) * sin (2pi*x/a) over the interval 0 to a.

    I have had no luck and can't find it in my tables. Do I have to convert it to exponentials and integrate all those terms?

    I tried using sin 2y = 2siny cosy to get
    [inte] y * 2sin2(y) cos(y) dy
    but that didn't get me anywhere. Help would be greatly appreciated.
     
    Last edited: Oct 25, 2003
  2. jcsd
  3. Oct 24, 2003 #2
    let u=siny
    then du=cosydy
    and [inte]2sin2y cosydy=2[inte]u2du=2/3u3
     
  4. Oct 24, 2003 #3
    That's not going to do it -- you overlooked the x term.

    But,
    xsin(x)sin(2x) = x*sin(x)*2sin(x)cos(x)
    = 2xsin2(x)cos(x)
    = 2x(1-cos2(x))*cos(x)
    = 2xcos(x) -2xcos3(x)

    and you can find
    ∫ 2xcos(x) dx
    and
    ∫ 2xcos3(x) dx
    using integration by parts.

    Can you take it from there?
     
  5. Oct 25, 2003 #4
    Re: Re: x * sin x * sin2x integral

    My fault, I left the y out of the integral you copied. But I did try this in combination with the y term and then integration by parts as
    u = y, du = dy and dv = your u substitution above. It didn't seem to work.
     
  6. Oct 25, 2003 #5
    I think I can I think I can ....

    The advantage of the conversion to cosines is that I get a sum of terms instead of a product of sines?
     
  7. Oct 25, 2003 #6

    Hurkyl

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    Let's take it from the top. Your application of a trigonometric identity was a good choice, because it makes all of the trig functions involved have the same argument.

    ∫ 2 y (sin y)2 cos y dy

    The others have mentioned integration by parts, and, well, this is a prototypical application of IBP since we have y multiplied by something we can integrate!

    So we do IBP, setting:

    u = 2y
    dv = (sin y)2 cos y dy

    du = 2 dy
    v = ∫ (sin y)2 cos y dy


    Stephen has given the standard approach on how to perform this subintegral. Applying the substitution will allow you to integrate and compute what v is supposed to be.

    Can you take it from here?
     
  8. Oct 27, 2003 #7
    is the integral of 2xcosx dx :

    2xsinx + cosx + c?


    and integral of 2x(cosx)^3 dx:

    xcos^4/2 - cos^5/20 + c?
     
  9. Oct 28, 2003 #8
    I got it thanks. Then I got lots of practice in variations of the same theme. Then I got to do lots of integrals with x2. These were much tougher.

    Does anyone know a book with higher powers of sin and cosine terms?

    I like Alan Jefferey's book but the tables stop at

    [inte] x sin3x dx and
    [inte] x2 sin2x dx

    In my problems I'm getting sin and cosines to the 4,5 and 6th powers.
     
  10. Oct 28, 2003 #9
    Here's what I got, 2 answers, don't know if any one of them is right.


    2[xsinx - cosx - 3xcosxsinx + 3cos^2(x) - 3sin^2(x)] + C

    where x = pi*x/a


    2(xsinx - cosx - 3xcosxsinx + 3cos^2(x) + sin^2(x)) + C

    where x = pi*x/a
     
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