Show using integration by parts that:
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
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.
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)
To solve the integration using integration by parts, you correctly identified the functions u and dv.
Let's continue with the solution:
u = e^3x
du/dx = 3e^3x
dv = sin(2x) dx
v = ∫ dv = ∫ sin(2x) dx = -1/2 cos(2x)
Now, we can use the formula for integration by parts:
∫ u dv = uv - ∫ v du
Plugging in the values:
∫ e^3x sin(2x) dx = -1/2 e^3x cos(2x) - ∫ (-1/2 cos(2x)) (3e^3x) dx
Now, we can simplify and rearrange the terms:
= -1/2 e^3x cos(2x) + 3/2 ∫ cos(2x) e^3x dx
Let's focus on the remaining integral:
∫ cos(2x) e^3x dx
To solve this, we can use integration by parts again. Following the same steps as before:
u = cos(2x)
du/dx = -2 sin(2x)
dv = e^3x dx
v = ∫ dv = ∫ e^3x dx = 1/3 e^3x
Now, we apply the integration by parts formula:
∫ cos(2x) e^3x dx = 1/3 e^3x cos(2x) - ∫ (1/3 e^3x) (-2 sin(2x)) dx
= 1/3 e^3x cos(2x) + 2/3 ∫ e^3x sin(2x) dx
Now, we can substitute this result back into the original equation:
∫ e^3x sin(2x) dx = -1/2 e^3x cos(2x) + 3/2 (1/3 e^3x cos(2x) + 2/3 ∫ e^3x sin(2x) dx)
Simplifying further:
∫ e^3x sin(2x) dx = -1/2 e^3x cos(2x) + 1/2 e^3x cos(2x) + 2/3 ∫ e^3x sin(2x) dx
The terms -1/2 e^3x cos(2x) cancel out:
∫ e^3x sin(2x) dx = 1/2 e^3x cos(2x) + 2/3 ∫ e^3x sin(2x) dx
Now, let's bring the remaining integral to one side:
∫ e^3x sin(2x) dx - 2/3 ∫ e^3x sin(2x) dx = 1/2 e^3x cos(2x)
Combining the integrals on the left side:
(1 - 2/3) ∫ e^3x sin(2x) dx = 1/2 e^3x cos(2x)
Simplifying:
1/3 ∫ e^3x sin(2x) dx = 1/2 e^3x cos(2x)
Now, let's solve for the integral alone:
∫ e^3x sin(2x) dx = (2/3)/(1/2) e^3x cos(2x)
= 4/3 e^3x cos(2x)
Therefore, the final answer is:
∫ e^3x sin(2x) dx = (4/3) e^3x cos(2x) + c
where c is the constant of integration.
I hope this explanation helps! Let me know if you have any further questions.