Joseph Krupa

2021-12-29

Use a table of integrals with forms involving eu to find the indefinite integral $\int {e}^{-4x}\mathrm{sin}3xdx$

braodagxj

Expert

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

lenkiklisg7

Expert

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

Vasquez

Expert

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