1. Integration

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

a) Differentiate $e^{-3x}sin(2x)$ and $e^{-3x}cos(2x)$ with respect to $x$.
$\frac{d}{dx}(e^{-3x}sin(2x))=e^{-3x}(2cos(2x)-3sin(2x))$
$\frac{d}{dx}(e^{-3x}cos(2x))=e^{-3x}(3cos(2x)+2sin(2x))$

b) Hence show that
$e^{-3x}sin(2x)+c_1 = -3{\int}e^{-3x}sin(2x)+2{\int}e^{-3x}cos(2x)$
and $e^{-3x}cos(2x)+c_2 = -3{\int}e^{-3x}cos(2x)-2{\int}e^{-3x}sin(2x)$
c) Use the two equations from b to determine ${\int}e^{-3x}sin(2x).$

2. Re: Integration

First of all, you have a sign error in \displaystyle \begin{align*} \frac{\mathrm{d}}{\mathrm{d}x} \left[ \mathrm{e}^{-3x}\cos{ \left( 2x \right) } \right] = -3\mathrm{e}^{-3x}\cos{ \left( 2x \right) } - 2\mathrm{e}^{-3x}\sin{ \left( 2x \right) } = -\mathrm{e}^{-3x} \left[ 3\cos{ \left( 2x \right) } + 2 \sin{ \left( 2x \right) } \right] \end{align*}. This might have affected your chances to get further...

3. Re: Integration

Thanks, but I'm still stuck on b and c.

4. Re: Integration

Nevermind, I worked it out.

5. Re: Integration

The answer for the second derivative of part (a) is missing a negative sign(-) in the front.

Part (b) follows from part (a) from the difference rule of integration.

Part (c) can be found by multiplying both equations from (b) by a constant so as to eliminate the ${\int}e^{-3x}cos(2x)$ when adding both equations together.