
#1
Oct2403, 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 * 2sin^{2}(y) cos(y) dy but that didn't get me anywhere. Help would be greatly appreciated. 



#2
Oct2403, 08:18 PM

P: 364

then du=cosydy and [inte]2sin^{2}y cosydy=2[inte]u^{2}du=2/3u^{3} 



#3
Oct2403, 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) = 2xsin^{2}(x)cos(x) = 2x(1cos^{2}(x))*cos(x) = 2xcos(x) 2xcos^{3}(x) and you can find ∫ 2xcos(x) dx and ∫ 2xcos^{3}(x) dx using integration by parts. Can you take it from there? 



#4
Oct2503, 01:44 AM

P: 211

x * sin x * sin2x integralu = y, du = dy and dv = your u substitution above. It didn't seem to work. 



#5
Oct2503, 01:48 AM

P: 211

The advantage of the conversion to cosines is that I get a sum of terms instead of a product of sines? 



#6
Oct2503, 12:30 PM

Emeritus
Sci Advisor
PF Gold
P: 16,101

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? 



#7
Oct2703, 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? 



#8
Oct2803, 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 x^{2}. 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 sin^{3}x dx and [inte] x^{2} sin^{2}x dx In my problems I'm getting sin and cosines to the 4,5 and 6th powers. 



#9
Oct2803, 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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