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Psyc
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[SOLVED] Integral of x^2*sin(ax)dx
So, I'm a newb at Integration, especially by parts.
This is what I have
[tex]\int x^{2}sin(ax)dx[/tex]
I set:
u=x[tex]^{2}[/tex] and dv=sin(ax)dx
so I got:
du=2xdx and v=-cos(ax)/a
So then I put everything together
[tex]u*v- \int v*du[/tex]
[tex]-x^{2}*\frac{-cos(ax)}{a} - \int \frac{-cos(ax)}{a}*2xdx[/tex]
[tex]-x^{2}*\frac{-cos(ax)}{a} + 2\int \frac{-cos(ax)}{a}*xdx[/tex]
So that's pretty much where I get lost. I don't know what do do about the second integral.
I tried setting u = 2x and dv = cos(ax)/a * dx and I got
-x^2*cos(ax)/a - 2xsin(ax) - 2cos(ax)
but WebAssign (website where I submit homework) says that's wrong.
This is my first post here. I had some trouble with some other problems that I have so I googled them and most of the search results pointed me here.
So, any suggestions are appreciated.
Thanks
So, I'm a newb at Integration, especially by parts.
This is what I have
[tex]\int x^{2}sin(ax)dx[/tex]
I set:
u=x[tex]^{2}[/tex] and dv=sin(ax)dx
so I got:
du=2xdx and v=-cos(ax)/a
So then I put everything together
[tex]u*v- \int v*du[/tex]
[tex]-x^{2}*\frac{-cos(ax)}{a} - \int \frac{-cos(ax)}{a}*2xdx[/tex]
[tex]-x^{2}*\frac{-cos(ax)}{a} + 2\int \frac{-cos(ax)}{a}*xdx[/tex]
So that's pretty much where I get lost. I don't know what do do about the second integral.
I tried setting u = 2x and dv = cos(ax)/a * dx and I got
-x^2*cos(ax)/a - 2xsin(ax) - 2cos(ax)
but WebAssign (website where I submit homework) says that's wrong.
This is my first post here. I had some trouble with some other problems that I have so I googled them and most of the search results pointed me here.
So, any suggestions are appreciated.
Thanks