How do you find the indefinite integral of #int e^(2x)sin(7x)dx#?

Question

Isaiah Allen · Accepted Answer

#1/53e^(2x)(2sin 7x-7cos 7x)+C#

Integration by parts:

#intudv=uv-intvdu#

#I=inte^(2x)sin 7xdx#

#e^(2x)=u => 2e^(2x)dx=du# #dv=sin 7xdx => v=intsin 7xdx=-1/7cos 7x#

#I=-1/7e^(2x)cos 7x+2/7inte^(2x)cos 7xdx#

Again:

#e^(2x)=u => 2e^(2x)dx=du#

#dv=cos 7xdx => v=intcos 7xdx=1/7sin 7x#

#I=-1/7e^(2x)cos 7x+2/7[1/7e^(2x)sin7x-2/7inte^(2x)sin7xdx]#

#I=-1/7e^(2x)cos 7x+2/49e^(2x)sin7x-4/49inte^(2x)sin7xdx#

#I=-1/7e^(2x)cos 7x+2/49e^(2x)sin7x-4/49I#

#I+4/49I=-1/7e^(2x)cos 7x+2/49e^(2x)sin7x#

#53/49I=-1/7e^(2x)cos 7x+2/49e^(2x)sin7x#

#I=-7/53e^(2x)cos 7x+2/53e^(2x)sin7x+C#

#I=1/53e^(2x)(2sin 7x-7cos 7x)+C#

Samantha Adkison · Answer

undefined To find the indefinite integral of $ \int e^{2x} \sin(7x) \, dx $, we can use integration by parts:

$$ \int u \, dv = uv - \int v \, du $$

Let:
$$ u = e^{2x} \quad \text{(choose this as the first function)} $$
$$ dv = \sin(7x) \, dx \quad \text{(this will be the second function)} $$

Now, differentiate $ u $ to find $ du $ and integrate $ dv $ to find $ v $:

$$ du = 2e^{2x} \, dx $$
$$ v = -\frac{1}{7}\cos(7x) $$

Apply the integration by parts formula:

$$ \int e^{2x} \sin(7x) \, dx = -\frac{1}{7} e^{2x} \cos(7x) - \int -\frac{1}{7}\cos(7x) \cdot 2e^{2x} \, dx $$

$$ = -\frac{1}{7} e^{2x} \cos(7x) + \frac{2}{7} \int e^{2x} \cos(7x) \, dx $$

Now, we have a new integral $ \int e^{2x} \cos(7x) \, dx $, which can be solved using integration by parts again or by using the method of undetermined coefficients.