Integral of x*sin(ax)


by ponjavic
Tags: integral, xsinax
ponjavic
ponjavic is offline
#1
Nov29-04, 06:46 AM
P: 226
What is the integral of x*sin(x) and x*sin(ax)?
I have no idea since I have neveer integrated something to get a product...
Ohh, it's supposed to be integrated from 0 to 1 for the sin(ax)
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HallsofIvy
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#2
Nov29-04, 07:01 AM
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Use "integration by parts".

From the product rule for derivatives, d(uv)/dx= u(dv/dx)+ v(du/dx). We can write that in "differential" form as d(uv)= u dv+ vdu and then rewrite it as

u dv= d(uv)- vdu.

Integrating both sides gives the integral formula
[tex]\int u dv= uv- \int vdu[/tex].

In particular, to integrate x sin(ax), let u= x, dv= sin(ax) dx. Then du= dx and
v= -(1/a)cos(ax) so
[tex]\int x sin(ax)dx= -(\frac{1}{a}x cos(ax)+ \frac{1}{a}\int cos(ax) dx[/tex]

[tex]= -\frac{1}{a}(x cos(x)+ \frac{1}{a}sin(ax))[/tex].
dextercioby
dextercioby is offline
#3
Nov29-04, 07:15 AM
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Quote Quote by HallsofIvy
Use "integration by parts".
In particular, to integrate x sin(ax), let u= x, dv= sin(ax) dx. Then du= dx and
v= -(1/a)cos(ax) so
[tex]\int x sin(ax)dx= -(\frac{1}{a}x cos(ax)+ \frac{1}{a}\int cos(ax) dx[/tex]

[tex]= -\frac{1}{a}(x cos(x)+ \frac{1}{a}sin(ax))[/tex].
Sorry,there's a minus,a paranthesis too much and an "a" missing:
[tex]\int x sin(ax)dx= -\frac{1}{a}x cos(ax)+ \frac{1}{a}\int cos(ax) dx[/tex]

[tex]= -\frac{1}{a}[x cos(ax)- \frac{1}{a}sin(ax)][/tex]

ponjavic
ponjavic is offline
#4
Nov29-04, 12:11 PM
P: 226

Integral of x*sin(ax)


and if its / ? =) as in sin(ax)/x
Or is it so easy that I can do it by myself, don't have time right now...
ponjavic
ponjavic is offline
#5
Nov29-04, 05:08 PM
P: 226
This is what I seem to get, very annoying
[tex]\int(sin(ax)\frac{1}{x})=sin(ax) ln(x)+\int(ln(x) cos(ax))[/tex]
or
[tex]\int(\frac{1}{x}sin(ax))=-\frac{1}{x}cos(ax)+\int(\frac{1}{x^2}cos(ax))[/tex]

Any ideas? Are any of the following integrals easy to do?
vladimir69
vladimir69 is offline
#6
Nov29-04, 05:34 PM
P: 131
did u try setting u=x and dv=sin(ax)dx ?
this is what i got
[tex] \int xsin(ax) dx=-\frac{x}{a}cos(ax) +\frac{1}{a}\int cos(ax) dx[/tex]
[tex] = -\frac{x}{a}cos(ax)+\frac{1}{a^2}sin(ax)[/tex]
ponjavic
ponjavic is offline
#7
Nov30-04, 06:41 AM
P: 226
heh vladimir, I understand the xsin(ax) integral but now I am trying to do 1/x*sin(ax) is this possible? If you look at my previous post you'll see me trying to integrate 1/x*sin(ax)
dextercioby
dextercioby is offline
#8
Nov30-04, 05:37 PM
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P: 11,863
That's because there is no primitive of the function sinx/x.
I assume you know that ordinary functions can be differentiated and the result be another "familiar" function.But this does not apply for primitives.There are functions like sinx/x,cos/x,exp(x^2),etc. which do not have primitives.That is,u cannot find a function which to differentiate to get the function you wish to integrate.
However,numerical methods based on Taylor/Mac Laurin formula(s) can be used to obtain results.For example,to find the primitive of sinx/x,u need to expand sinx and devide each term of the expansion term by x and integrate the results.You'll have then a new infinite series,which could be seen as the Taylor/Mac Laurin exapansion of the function u are looking for.
This thing works for functions which "behave" pretty well as to apply Taylor/Mac Laurin formula(s) to them.The 3 examples i have stated prove this assertion.
To find definite integral values for the 3 functions mentioned above,try to get a hand on 2 books:M.Abramowitz,I.Segun:"Mathematical functions and tables" and Rytzhik and Gradstein:"Tables of integrals" and search for sine integral function,cosine integral function and erf(error) function.

P.S.I'm not at the library anymore,so from now on,when i give indications to certain books always doubt the veridicity of the names and titles stated,as i give them from my memory to which i have no recollection of having ever been treated with glucosis.So it cold fail me someday.Hopefully not soon.
ponjavic
ponjavic is offline
#9
Dec1-04, 03:28 AM
P: 226
Thanks man, I was expecting the likes...


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