Working from the general form integral

1. The problem statement, all variables and given/known data
lot of work to go around here so need to see if i did it right. thanks!
[tex]\int e^2xsin3xdx[/tex]

2. Relevant equations
[tex]\int uav=uv-\int vdu[/tex]


3. The attempt at a solution
[tex]\int\frac{-e^{2x}}{3}cos3x+\frac{2}{3}\int(e^{2x})cos3xdx[/tex]

parts work: let u=e^2x, dv= sin3xdx
so: du=2e^2x, and v= [tex]\frac{-1}{3}cos3x[/tex]

next:[tex]\int(e^{2x}cos3xdx)=\frac{e^{2x}}{3}sin3x-\frac{-2}{3}\int(e^{2x}sin3xdx)[/tex]

work:let u=[tex]\frac{-e^{2x}}{3}, dv= cos3xdx, du= 2e^{2x} v= \frac{1}{3}sin3x[/tex]

now: [tex]\int e^{2x}sin3xdx= \frac{-e^{2x}}{3}cos3x+\frac{2}{3}[\frac{e^{2x}}{3}sin3x-\frac{2}{3}\int e^{2x}sin3xdx][/tex]

" " "= [tex]\int e^{2x}sin3xdx+ \frac{-4}{9} \int e^{2x}sin3xdx= \frac{-e^{2x}}{3}cos3x+\frac{2}{9}e^{2x}sin3x[/tex]

[tex]\frac{9}{13}\frac{13}{9}\int e^{2x}sin3xdx= [\frac{-e^{2x}}{3}cos3x+\frac{2}{9}e^{2x}sin3x]\frac{9}{13}[/tex]

[tex]\int e^{2x}sin3xdx = \frac{-3}{13}e^{2x}cos3x+ \frac{2}{13}e^{2x}sin3x+C = e^{2x} [ \frac{2}{13}sin3x- \frac{3}{13}cos3x+C[/tex]

ok now last: a=2 b=3

[tex]\int e^{2x}sin3xdx= \frac{1}{2^{2}+3^{2}}e^{2x}[2sin3x-3cos3x]+c= e^{2x}[\frac{2}{13}sin3x-\frac{3}{13}cos3x]+C[/tex]

done. i think. we learned this a few days back and was given it to see if we can do it. it is extra work so it wont kill my grade or anything for getting it wrong, but it will help me if i can get it right. i looked it over and to me it looks fine just want to see if anyone can see if i overlooked something.

also i hope it looks right this is the 1st time i am using this sight(well used it few days back but not for anything like this)