Use a table of integrals with forms involving eu to find the indefinite integral ∫e−4xsin⁡3xdx

Step 1
By using the table of integrals we are to find the indefinite integral:
${\displaystyle \int e^{-4x}\sin 3xdx}$
Step 2
We will use formula from the table of integrals with forms involving ${\displaystyle e^u}$:
${\displaystyle \int e^{ux}\sin vxdx=\frac{e^{ux}}{u^2+v^2}[u\sin vx-v\cos vx]+C}$...(1)
We are to find the indefinite integral ${\displaystyle \int e^{-4x}\sin 3xdx}$
So, Here u=-4 and v=3
On Putting u=-4 and v=3 in (1) we get:
${\displaystyle \int e^{-4x}\sin 3xdx=\frac{e^{-4x}}{{(-4)}^2+{\left(3\right)}^2}[-4\sin 3x-3\cos 3x]+C}$
${\displaystyle \int e^{-4x}\sin 3xdx=\frac{e^{-4x}}{25}[-4\sin 3x-3\cos 3x]+C}$
or ${\displaystyle \int e^{-4x}\sin 3xdx=-\frac{e^{-4x}}{25}[4\sin 3x+3\cos 3x]+C}$

${\displaystyle \int e^{-4x}\sin \left(3x\right)dx}$
The formula for integration by parts:
${\displaystyle \int U\left(x\right)\cdot dV\left(x\right)=U\left(x\right)\cdot V\left(x\right)-\int V\left(x\right)\cdot dU\left(x\right)}$
We put
U=1
dV=dx
Then:
dU=0dx
V=x
Therefore:
${\displaystyle \int e^{-4x}\cdot \sin \left(3x\right)dx=x}$
Answer:
${\displaystyle \int e^{-4x}\cdot \sin \left(3x\right)=x+C}$

$\begin{array}{}\int e^{-4x}\sin (3x)dx\\ =-\frac{e^{-4x}\sin (3x)}{4}-\int -\frac{3e^{-4x}\cos (3x)}{4}dx\\ =-\frac{e^{-4x}\sin (3x)}{4}-(\frac{3e^{-4x}\cos (3x)}{16}-\int -\frac{9e^{-4x}\sin (3x)}{16}dx)\\ =-\frac{e^{-4x}\sin (3x)}{4}-(\frac{3e^{-4x}\cos (3x)}{16}+\frac{9}{16}\int e^{-4x}\sin (3x)dx)\\ =\frac{-4e^{-4x}\sin (3x)-3e^{-4x}\cos (3x)}{25}\\ \int e^{-4x}\sin (3x)dx\\ =\frac{-4e^{-4x}\sin (3x)-3e^{-4x}\cos (3x)}{25}+C\\ =-\frac{e^{-4x}(4\sin (3x)+3\cos (3x))}{25}+C\end{array}$