# x * sin x * sin2x integral

by mmwave
Tags: integral, sin2x
 P: 211 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.
P: 364
 Originally posted by mmwave I tried using sin 2y = 2siny cosy to get [inte] 2sin^2 (y) cos (y) dy but that didn't get me anywhere. Help would be greatly appreciated.
let u=siny
then du=cosydy
and [inte]2sin2y cosydy=2[inte]u2du=2/3u3
 P: 1,048 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?
P: 211

## x * sin x * sin2x integral

 Originally posted by StephenPrivitera let u=siny then du=cosydy and [inte]2sin2y cosydy=2[inte]u2du=2/3u3
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.
P: 211
 Originally posted by gnome 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?
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?
 PF Patron Sci Advisor Emeritus P: 16,094 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?
 P: 640 is the integral of 2xcosx dx : 2xsinx + cosx + c? and integral of 2x(cosx)^3 dx: xcos^4/2 - cos^5/20 + c?
 P: 211 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.
 P: 640 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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