To solve this integral, we can use the method of integration by parts. The formula for integration by parts is given by:
∫udv = uv - ∫vdu
First, we need to identify the u and the dv part from our integral. We need to consider the easier part to differentiate as u and the easier part to integrate as dv. Here, u = e⁻²ˣ and dv = cos(3x)dx.
Now, we take the derivative of u, du = -2e⁻²ˣdx and the integral of dv, v = sin(3x)/3.
Substituting u, v, du, dv in the integration by parts formula, we get:
∫e⁻²ˣcos(3x)dx = e⁻²ˣ*sin(3x)/3 - ∫(-2e⁻²ˣ*sin(3x)/3)dx
We see that the integral on the right-hand side is still not easily solvable, so we apply integration by parts once again on that. This time, let u = -2e⁻²ˣ/3 and dv = sin(3x)dx.
Therefore, du = (4/3)e⁻²ˣdx and v = -cos(3x)/3.
Plugging the new u, v, du, dv into the integration by parts formula, we get:
∫(-2e⁻²ˣ*sin(3x)/3)dx = -2e⁻²ˣ*cos(3x)/9 - ∫(-4/9)e⁻²ˣcos(3x)dx
We add this into our above equation:
∫e⁻²ˣcos(3x)dx = e⁻²ˣ*sin(3x)/3 + 2e⁻²ˣ*cos(3x)/9 + ∫(4/9)e⁻²ˣcos(3x)dx
We observe that the integral on both sides is the same, hence we proceed to solve the equation for the integral:
Let I = ∫e⁻²ˣcos(3x)dx
Subtracting I from both sides, we get:
8/9*I = e⁻²ˣ*sin(3x)/3 + 2e⁻²ˣ*cos(3x)/9
Solving for I, we get:
I = [3e⁻²ˣ*sin(3x)/13 - 2e⁻²ˣ*cos(3x)/13]
So, ∫e⁻²ˣcos(3x)dx = [3e⁻²ˣ*sin(3x)/13 - 2e⁻²ˣ*cos(3x)/13].