The following function is being integrated:
$$e^{-3x}\sin(3x)$$
For the ease of time, I have managed to calculate it down to the following formula using integration by parts:
$$0=-e^{-3x}\cos(3x)+e^{-3x}\sin(3x)$$
The above phrase is calculated by splitting the first phrase into two.
Where$$u=e^{-3x}$$and$$v=\sin(3x)$$
Would this be an appropriate solution in terms of its simplification and finding the integral of the initial equation? Can I go and do anything further?
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$\begingroup$Here's a fun way to do it if you're familiar with complex numbers: Let$$I=\int e^{-3x}\sin(3x)dx=\int e^{-3x}\Im e^{3ix}dx=\tag1$$ $$\Im\int e^{-3x+3ix}dx=\Im \int e^{x(-3+3i)}dx=\Im\left(\frac{1}{-3+3i}e^{-3x+3ix}\right)\tag2$$ $$=\Im\left(e^{-3x}\left(-\frac{1}{6}-\frac{i}{6}\right)\left(\cos(3x)+i\sin(3x)\right)\right)\tag3$$ $$=-\frac{e^{-3x}}{6}\Im\left(-\sin(3x)+\cos(3x)+i\left(\sin(3x)+\cos(3x)\right)\right)\tag4$$ $$\boxed{=-\frac{e^{-3x}}{6}(\sin(3x)+\cos(3x))+C}\tag5$$
Alternatively, integrate by parts twice making sure to keep track of your integral by naming it the 'variable' $I$; namely
Let $$I=\int e^{-3x}\sin(3x)dx$$
Integrate by parts, $$u=e^{-3x}\implies du=-3e^{-3x}\space dx$$ $$dv=\sin(3x)dx\implies v=-\frac{1}{3}\cos(3x)$$ Then $$I=-\frac{1}{3} e^{-3x}\cos(3x)-\int e^{-3x}\cos(3x)dx $$
Integrate by parts one more time on the last integral, $$t=e^{-3x}\implies dt=-3e^{-3x}\space dx$$ $$dw=\cos(3x)dx\implies w=\frac{1}{3}\sin(3x)$$
so that $$I=-\frac{1}{3} e^{-3x}\cos(3x)-\left(\frac{1}{3}e^{-3x}\sin(3x)-\int-e^{-3x}\sin(x)dx\right) $$ $$=-\frac{e^{-3x}}{3}\left(\sin(3x)+\cos(3x)\right)-I\implies2I=-\frac{e^{-3x}}{3}\left(\sin(3x)+\cos(3x)\right)$$ So dividing by $2$ on both sides gives $$I=-\frac{e^{-3x}}{6}\left(\sin(3x)+\cos(3x)\right)+C$$ matching the above formula. $$$$
$(1)$: Utilize Euler's formula $e^{3ix}=\cos(3x)+i\sin(3x)$ and $\Im(e^{3ix})=\sin(3x)$
$(2)$: Utilize the facts: $e^ae^b=e^{a+b}$ and $\int e^{kx}dx=\frac{1}{k}e^{kx}+c$
$(3)$: Simplify $\frac{1}{-3+3i}$ and expand $e^{-3x+3ix}=e^{-3x}e^{3ix}=e^{-3x}(\cos(3x)+i\sin(3x))$
$(4)$: Expand $\left(-\frac{1}{6}-\frac{i}{6}\right)\left(\cos(3x)+i\sin(3x)\right)$ and collect real and imaginary parts
$(5)$: Take the imaginary part
$\endgroup$ $\begingroup$In general the formula for integration by parts is $\int uv'=uv-\int u'v$.
If $u = e^{-3x}$, then $u'=-3e^{-3x}$ and if $v' = \sin(3x)$, then $v = -\frac{1}{3}\cos(3x)$.
So $\int e^{-3x} \sin(3x)dx = -\frac{1}{3}e^{-3x}\cos(3x)-\color{red}{\int e^{-3x} \cos(3x)dx} + C$.
If $u = e^{-3x}$, then $u'=-3e^{-3x}$ and if $v' = \cos(3x)$, then $v = \frac{1}{3}\sin(3x)$.
So $\color{red}{\int e^{-3x} \cos(3x)dx} = \color{blue}{\frac{1}{3}e^{-3x}\sin(3x)+\int e^{-3x}\sin(3x)} + C'$.
Plug in the result for the red part in the equation above and you get:
$\int e^{-3x} \sin(3x)dx = -\frac{1}{3}e^{-3x}\cos(3x)-\left(\color{blue}{\frac{1}{3}e^{-3x}\sin(3x)+\int e^{-3x}\sin(3x)}\right)+C''$
Solve this equation for $\int e^{-3x} \sin(3x)dx$ and you get this:
$\int e^{-3x} \sin(3x)dx = -\frac{1}{6}e^{-3x}(\cos(3x)+\sin(3x))+C''$
$\endgroup$ $\begingroup$You can do this way using two integrations by part. The idea is to integrate the $e^{-3x}$ part twice and to recognize the original integral after the second integration by part.
\begin{align*}I&=\int e^{-3x}\sin(3x) dx=\left[-\dfrac{e^{-3x}}{3} \sin(3x)\right]+\int_a^b {e^{-3x}}\cos(3x)dx\\ &=\left[-\dfrac{e^{-3x}}{3} \sin(3x)\right] + \left[-\dfrac{e^{-3x}}{3} \cos(3x)\right] - \int {e^{-3x}}\sin(3x)dx\\ 2I&=-\dfrac{e^{-3x}}{3}\left(\sin(3x)+\cos(3x)\right)+C \end{align*}Hence $$I=-\dfrac{e^{-3x}}{6}\left(\sin(3x)+\cos(3x)\right)+C.$$
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