I'm having trouble with this rather simple problem while revising integration:

a) Differentiate $\displaystyle e^{-3x}sin(2x)$ and $\displaystyle e^{-3x}cos(2x)$ with respect to $\displaystyle x$.

$\displaystyle \frac{d}{dx}(e^{-3x}sin(2x))=e^{-3x}(2cos(2x)-3sin(2x))$

$\displaystyle \frac{d}{dx}(e^{-3x}cos(2x))=e^{-3x}(3cos(2x)+2sin(2x))$

b)Hence show that

$\displaystyle e^{-3x}sin(2x)+c_1 = -3{\int}e^{-3x}sin(2x)+2{\int}e^{-3x}cos(2x)$

and $\displaystyle e^{-3x}cos(2x)+c_2 = -3{\int}e^{-3x}cos(2x)-2{\int}e^{-3x}sin(2x)$

c)Use the two equations frombto determine $\displaystyle {\int}e^{-3x}sin(2x).$

Thank you for your time.