Let M = ∫ e^(2x) cos3x dx
u = cos3x
du = -3sin3x dx
dv = e^(2x) dx
v = 1/2 e^(2x)
M = uv - ∫v du
M = 1/2 e^(2x) cos3x + (3/2)∫e^(2x) sin3x dx
Now consider ∫e^(2x) sin3x dx
u = sin3x
du = 3cos3x dx
dv = e^(2x) dx
v = 1/2 e^(2x)
∫e^(2x) sin3x dx = 1/2 e^(2x) sin3x - 3/2 ∫e^(2x) cos3x dx
= 1/2 e^(2x) sin3x - 3/2 M
So, now we have
M = 1/2 e^(2x) cos3x + (3/2)[1/2 e^(2x) sin3x - 3/2 M]
M = 1/2 e^(2x) cos3x + 3/4 e^(2x) sin3x - 9/4 M
13/4 M = 1/4 e^(2x) (2cos3x+3sin3x)
M = 1/13 e^(2x) (2cos3x+3sin3x) + C
Note that you could just as well have used
u = e^(2x)
du = 2e^(2x) dx
dv = cos3x dx
v = 1/3 sin3x
and again two integrations by parts would have worked.
Not sure how this would have worked using the tabular method, since there is no nth derivative which is zero.
Evaluate using tabular method:
integral e^(2x)cos3xdx
1 answer