How do you integrate #int e^(5x)cos3x#?

Question

Luke Alvarez · Accepted Answer

You can integrate by parts twice and algebraically solve for the integral to get #int\ e^{5x}cos 3x\ dx=5/34 e^{5x}cos 3x+3/34 e^{5x}sin 3x+C# Let #I=int\ e^{5x}cos 3x\ dx#. Now let #u=e^{5x}# and #dv=cos 3x\ dx# so that #du=5e^{5x}\ dx# and #v=1/3 sin 3x#. Then #I=1/3 e^{5x}sin 3x-5/3int\ e^{5x}sin 3x\ dx#. For this next integral, let #u=e^{5x}# and #dv=sin 3x\ dx# so that #du=5e^{5x}# and #v=-1/3 cos 3x#. It follows that #I=1/3 e^{5x}sin 3x-5/3(-1/3 e^{5x}cos 3x+5/3 int e^{5x}cos 3x\ dx)# #=1/3 e^{5x}sin 3x+5/9e^{5x}cos 3x-25/9 I#. Therefore, #34/9 I=1/3 e^{5x}sin 3x+5/9 e^{5x}cos 3x#. Multiplying both sides by #9/34#, rearranging, and tacking on a #+C# at the end gives #I=int\ e^{5x}cos 3x\ dx=5/34 e^{5x}cos 3x+3/34 e^{5x}sin 3x+C#.

Scarlett Allison · Answer

undefined To integrate $ \int e^{5x}\cos(3x) $, you can use integration by parts. Let $ u = e^{5x} $ and $ dv = \cos(3x) dx $. Then, $ du = 5e^{5x} dx $ and $ v = \frac{1}{3} \sin(3x) $. Applying the integration by parts formula $ \int u \, dv = uv - \int v \, du $, you'll get:

$$ \int e^{5x}\cos(3x) \, dx = \frac{1}{3} e^{5x} \sin(3x) - \frac{5}{3} \int e^{5x} \sin(3x) \, dx $$

This integral on the right side can be further solved by another integration by parts or by using other techniques like integration of trigonometric functions with exponential functions.