I need to find the integral of e^(2x)sin(3x)
I used integration by parts and I let u=e^2x and dv=sin(3x)
My final answer was (-3/8)(e^(2x))(cos(3x)) - (1/8)(e^(2x))(sin(3x)) but it's wrong. Please help!!
u = e^(2x)
du/dx = 2 e^(2x)
du = 2 e^(2x) dx
dv = sin(3x) dx
v = -(1/3) cos(3x)
∫e^(2x)sin(3x) dx
= (e^(2x) )(-1/3)cos(3x) - ∫(-2/3)(e^(2x))(cos(3x) ) dx
=(e^(2x) )(-1/3)cos(3x) + (2/3)∫( e^(2x) (cos(3x)) dx
do it again on that last part:
let u = e^2x
du/dx = 2e^(2x)
let dv = cos(3x) dx
v = (1/3)sin(3x)
and ∫( e^(2x) (cos(3x)) dx
= (1/3)e^(2x) (sin(3x)) - ∫(1/3)sin(3x) (2e^(2x)) dx
= (1/3)e^(2x) (sin(3x)) - (2/3) ∫e^(2x) sin(3x) dx
but look at that last part, isn't that what we started with on the original left side of our equation
let ∫e^(2x)sin(3x) dx = A
so we have
A = (e^(2x) )(-1/3)cos(3x) + (2/3)[(1/3)e^(2x) (sin(3x)) - (2/3) A ]
A = (-1/3)(e^(2x)) (cos(3x) + (2/9)e^(2x) sin(3x) - (4/9)A
times 9
9A = -3 e^(2x) cos(3x) + 2 e^(2x) sin(3x) - 4A
13A = -3 e^(2x) cos(3x) + 2 e^(2x) sin(3x)
A = (-1/13) ( 3 e^(2x)cos(3x) - 2 e^(2x) sin(3x)
∫e^(2x)sin(3x) dx = (-1/13) ( 3 e^(2x)cos(3x) - 2 e^(2x) sin(3x) )
or as Wolfram has it
(1/13) ( 2 e^(2x) sin(3x) - 3 e^(2x)cos(3x) )
http://integrals.wolfram.com/index.jsp?expr=e%5E%282x%29sin%283x%29+&random=false
Thank you so much for showing the steps, I really appreciate the help! (:
To find the integral of e^(2x)sin(3x), you correctly applied the method of integration by parts. However, it seems like you made a mistake in differentiating u=e^(2x) and integrating dv=sin(3x).
Let's go through the correct steps of integration by parts:
Step 1: Choose u and dv.
In this case, let u = e^(2x) and dv = sin(3x).
Step 2: Find du and v.
To find du, differentiate u with respect to x: du/dx = 2e^(2x).
To find v, integrate dv with respect to x: v = (-cos(3x))/3.
Step 3: Apply the integration by parts formula.
The formula for integration by parts is:
∫ u dv = uv - ∫ v du
Step 4: Calculate the integral.
Using the integration by parts formula, we have:
∫ e^(2x)sin(3x) dx = u v - ∫ v du
= e^(2x) * (-cos(3x))/3 - ∫ (-cos(3x))/3 * 2e^(2x) dx
Simplifying the last term, we have:
= (-1/3)e^(2x)cos(3x) - 2/3 ∫ e^(2x)cos(3x) dx
Now, we have encountered another integral. We'll need to evaluate this second integral using integration by parts again. Let's repeat the process:
Step 1: Choose u and dv.
In this case, let u = e^(2x) and dv = cos(3x).
Step 2: Find du and v.
To find du, differentiate u with respect to x: du/dx = 2e^(2x).
To find v, integrate dv with respect to x: v = sin(3x)/3.
Step 3: Apply the integration by parts formula.
Using the formula ∫ u dv = uv - ∫ v du, we have:
∫ e^(2x)cos(3x) dx = u v - ∫ v du
= e^(2x) * sin(3x)/3 - ∫ (sin(3x)/3) * 2e^(2x) dx
= e^(2x) * sin(3x)/3 - (2/3) ∫ e^(2x)sin(3x) dx
Step 4: Rearrange the equation to solve for the original integral.
Let's substitute this back into the previous equation and solve for the original integral:
∫ e^(2x)sin(3x) dx = (-1/3)e^(2x)cos(3x) - (2/3) * [e^(2x) * sin(3x)/3 - (2/3) ∫ e^(2x)sin(3x) dx]
Now, let's move the integral term to one side of the equation:
∫ e^(2x)sin(3x) dx + (2/3) ∫ e^(2x)sin(3x) dx = (-1/3)e^(2x)cos(3x) - (2/9) e^(2x) * sin(3x)
Combining the integral terms:
(5/3) ∫ e^(2x)sin(3x) dx = (-1/3)e^(2x)cos(3x) - (2/9) e^(2x) * sin(3x)
Finally, solving for the integral:
∫ e^(2x)sin(3x) dx = [(-1/3)e^(2x)cos(3x) - (2/9) e^(2x) * sin(3x)] / (5/3)
Simplifying further, we have:
∫ e^(2x)sin(3x) dx = (-3/5) e^(2x)cos(3x) - (2/15) e^(2x) * sin(3x)
So, your final answer should be (-3/5) e^(2x)cos(3x) - (2/15) e^(2x) * sin(3x).