How do you find the integral of #e^(2x) cos4x dx#?

Question

Paisley Johnson · Accepted Answer

Integrate by parts twice using #u=e^(2x)# both times.

#I = int e^(2x) cos4x dx# #u= e^(2x)# and #dv = cos4x dx# #I = uv-intvdu # #= 1/4e^(2x)sin4x-1/2inte^(2x)sin4xdx# Now take #u = e^(2x)# and #dv = sin4xdx# #I=1/4e^(2x)sin4x-1/2[-1/4e^(2x)cos4x +1/2inte^(2x)cos4xdx]# #I = 1/4e^(2x)sin4x+ 1/8 e^(2x)cos4x - 1/4 I# Solve for #I = 1/5e^(2x)sin4x+ 1/10 e^(2x)cos4x +C#

William Abdalla · Answer

undefined To find the integral of $ e^{2x} \cos(4x) \, dx $, you can use integration by parts. Let $ u = e^{2x} $ and $ dv = \cos(4x) \, dx $. Then, $ du = 2e^{2x} \, dx $ and $ v = \frac{1}{4} \sin(4x) $.

Applying the integration by parts formula $ \int u \, dv = uv - \int v \, du $, we have:

$$ \int e^{2x} \cos(4x) \, dx = \frac{1}{4} e^{2x} \sin(4x) - \frac{1}{2} \int e^{2x} \sin(4x) \, dx $$

Now, we have another integral of the form $ e^{2x} \sin(4x) $. To solve this, we can use integration by parts again. Let $ u = e^{2x} $ and $ dv = \sin(4x) \, dx $. Then, $ du = 2e^{2x} \, dx $ and $ v = -\frac{1}{4} \cos(4x) $.

Applying the integration by parts formula again, we get:

$$ \int e^{2x} \sin(4x) \, dx = -\frac{1}{4} e^{2x} \cos(4x) - \frac{1}{2} \int e^{2x} \cos(4x) \, dx $$

Substituting this result back into our original integral, we have:

$$ \int e^{2x} \cos(4x) \, dx = \frac{1}{4} e^{2x} \sin(4x) + \frac{1}{8} e^{2x} \cos(4x) - \frac{1}{4} \int e^{2x} \cos(4x) \, dx $$

Now, we can solve for the integral by isolating it on one side:

$$ \frac{5}{4} \int e^{2x} \cos(4x) \, dx = \frac{1}{4} e^{2x} \sin(4x) + \frac{1}{8} e^{2x} \cos(4x) $$

$$ \int e^{2x} \cos(4x) \, dx = \frac{1}{5} e^{2x} \sin(4x) + \frac{1}{10} e^{2x} \cos(4x) + C $$

Where $ C $ is the constant of integration.