Show using integration by parts that:

Question

e^3x sin(2x)dx =

4/26 e^3x (3/2 sin(2x) - cos(2x)) +c

Bit stuck on this.

Using rule f udv = uv - f vdu

u = e^3x
dv + sin(2x)dx
f dv = v

du/dx = 3e^3x
v = -1/2 cos(2x)

so uv - f vdu: =

(e^3x)(-1/2 cos(2x)) - (-1/2 cos(2x))(3e^3x)

Don't know where to go or get to the final answer?? Any help greatly appreciated

Answer 1

In google type:

integration by parts online emathelp

When you see list of results cilck on:

Integral (Antiderivative) Calculator with Steps - eMathHelp

When page be open in rectangle type:

e^(3x) sin (2x)

then clic option:

CALCULATE

You wil see solution step-by-step.

Answer 2

I agree with what you have so far, other than skipping the integral sign and the dx from the last term.
You seem to have the concept however.

∫e^(3x) sin(2x) dx
= (-1/2)e^(3x)cos(2x) - ∫(-1/2)(3e^(3x))(cos(2x) dx
= (-1/2)e^(3x)cos(2x) + (3/2)∫ e^(3x) cos(2x) dx

now look at the ∫e^(3x) cos(2x) dx
at the end. Isn't that basically the same pattern as you just did?

so let's repeat:
let u = e^3x
du = 3 e^(3x) dx

let dv = cos(2x) dx
v = (1/2)sin(2x)

∫e^(3x) cos(2x) dx
= e^(3x)(1/2)sin(2x) - ∫(1/2)sin(2x)(3 e^(3x)) dx
= (1/2) e^(3x)sin(2x) - (3/2)∫ e^(3x)sin(2x) dx

ahhhh, but isn't ∫ e^(3x)sin(2x) dx what we started out with at the beginning

so if we let
k = ∫e^(3x) sin(2x) dx , for easier typing , then

k = (-1/2)e^(3x)cos(2x) + (3/2)∫ e^(3x) cos(2x) dx
k = (-1/2)e^(3x)cos(2x) + (3/2) [(1/2) e^(3x)sin(2x) - (3/2)∫ e^(3x)sin(2x) dx]
k = (-1/2)e^(3x)cos(2x) + (3/2) [(1/2) e^(3x)sin(2x) - (3/2) k ]

k = (-1/2)e^(3x)cos(2x) + (3/4) e^(3x)sin(2x) - (9/4) k
times 4
4k = -2e^(3x)cos(2x) + 3 e^(3x)sin(2x) - 9k
13k = e^(3x) (3sin(2x) -2cos(2x)

k = (e^(3x) (3sin(2x) -2cos(2x))/13

∫e^(3x) sin(2x) dx = (e^(3x) (3sin(2x) -2cos(2x))/13

as confirmed by Wolfram:
http://www.wolframalpha.com/input/?i=integral+e%5E(3x)+sin(2x)