Sending to my school's Second in Maths a corrected solution to an old S'level Maths question:
Integrate the function: x.e^ax. sinbx
???x e^ax sinbx dx=??x e^ax Im[e^ibx ] ? dx=Im[???x e^ax e^ibx ? dx]=Im[???x e^(a+ib)x dx?]
???x e^(a+ib)x dx?= [x ?(1/(a+ib) e^(a+ib)x )]- ???(1/(a+ib) e^(a+ib)x ) dx
= (?(1/(a+ib)))x e^(a+ib)x- (?(1/(a+ib)))^2 e^(a+ib)x
= (?((a-ib)/(a^2+b^2 )))x e^(a+ib)x- ?(?(a-ib)?^2/(a^2+b^2 )^2 ) e^(a+ib)x
We need the Imaginary part of this:
(?(a/(a^2+b^2 )))x e^ax sinbx-(?(b/(a^2+b^2 )))x e^ax cosbx- ?((a^2-b^2)/(a^2+b^2 )^2 ) e^ax sinbx- ?((-2ab)/(a^2+b^2 )^2 ) e^ax cosbx
= (?((a^3 x+ab^2 x-a^2+b^2)/(a^2+b^2 )^2 )) e^ax sinbx+(?((b(a^2 x+b^2 x-2a))/(a^2+b^2 )^2 )) e^ax cosbx
=?(e^ax/(a^2+b^2 )^2 ) [(a^3 x+ab^2 x-a^2+b^2)sinbx-b(a^2 x+b^2 x-2a)cosbx]
Something like that, anyway! No, that is now correct!
(You'll have to imagine integral signs and square brackets where the ? marks are, as appropriate)
One of my more accessible posts!