x * sin x * sin2x integral


by mmwave
Tags: integral, sin2x
mmwave
mmwave is offline
#1
Oct24-03, 07:45 PM
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.
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StephenPrivitera
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#2
Oct24-03, 08:18 PM
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
gnome
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#3
Oct24-03, 08:40 PM
P: 1,047
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?

mmwave
mmwave is offline
#4
Oct25-03, 01:44 AM
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.
mmwave
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#5
Oct25-03, 01:48 AM
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?
Hurkyl
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#6
Oct25-03, 12:30 PM
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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?
PrudensOptimus
PrudensOptimus is offline
#7
Oct27-03, 08:28 PM
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?
mmwave
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#8
Oct28-03, 02:06 PM
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.
PrudensOptimus
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#9
Oct28-03, 06:20 PM
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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