V_R = V_in + V_C (since voltage at a junction should be
single-valued)
=> -iR = sin(wt) + (1/C) (int from -infty to t) i(t') dt'
differentiate the above w.r.t. t :
-(di/dt) R = w cos(wt) + (1/C) i
take Laplace transform here, multiplying both sides by exp(-st)
then integrating from 0 to infty:
-Rs L(s) = -s/(s^2+w^2) + (1/C) L(s) where L(s) is the
Laplace transform of i(t)
rearrange to obtain
L(s)=( s/R ) ( s^3 + s^2/RC + sw^2 + w^2/RC )
do the usual partial fraction procedure:
L(s)= (1/R) [ (alpha s + beta) / ( s^2 + w^2 ) - alpha / (
s + 1/RC ) ]
where alpha and beta are coefficients I'm too lazy to specify
here
Now do the inverse Laplace transform by referring to tables (or
if you are masochistic enough, you can do the contour integrals)
and you will obtain i(t).
