Question

Evaluate using tabular method:

integral e^(2x)cos3xdx

Answer 1

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.