# X * sin x * sin2x integral

mmwave
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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## Answers and Replies

StephenPrivitera
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
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
&int; 2xcos(x) dx
and
&int; 2xcos3(x) dx
using integration by parts.

Can you take it from there?

mmwave

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
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
&int; 2xcos(x) dx
and
&int; 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?

Staff Emeritus
Gold Member
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.

&int; 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 = &int; (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
is the integral of 2xcosx dx :

2xsinx + cosx + c?

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

xcos^4/2 - cos^5/20 + c?

mmwave
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
Here's what I got, 2 answers, don't know if anyone 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