How do I integrate e^2x cos5x dx?

Answer 1

integrates by parts:

U = cos(5x) and dV = exp(2x)

dU = -5(sin(5x)) and V = (1/2)exp(2x)

UV - integral ( V*dU) =

(1/2) * cos(5x)*exp(2x) + (5/2) * integral ( exp(2x) * sin(5x) )

Integrating by parts again in the same way:

U = sin(5x) and dV = exp(2x)

dU = 5*cos(5x) V = (1/2) exp(2x)

This gives the same integral as posed in the problem, as highlighted in bold above

(1/2) * cos(5x)*exp(2x) + (5/2) * [ (1/2) sin(5x)*exp(2x) - (5/2) integral ( cos(5x)*exp(2x))]

(1/2) * cos(5x) * exp(2x) + (5/4) * sin(5x)*exp(2x) - 25/4 * integral( cos(5x)*exp(2x)]

solving for the integral, the solution is:

(4/29) [ (1/2) * cos(5x) * exp(2x) + (5/4) * sin(5x)*exp(2x) ]

(2/29) cos(5x) * exp(2x) + (5/29) sin(5x)*exp(2x) =

exp(2x) [2*cos(5x) + 5*sin(5x)]/29

Answer 2

Use integration by parts:

Let u = e^2x and dv = cos(5x)dx

Then, du = 2e^2xdx and v = (1/5)sin(5x)

so, ∫e^2xcos(5x)dx = (1/5)e^2xsin(5x) - (2/5)∫e^2xsin(5x)dx

Use integration by parts again with u = e^2x and dv = sin(5x)dx

So, du = 2e^2xdx and v = (-1/5)cos(5x)

We then have ∫e^2xcos(5x)dx = (1/5)e^2xsin(5x) - (2/5)[(-1/5)e^2xcos(5x) + (2/5)∫e^2xcos(5x)dx]

So, (29/25)∫e^2xcos(5x)dx = (1/5)e^2xsin(5x) + (2/25)e^2xcos(5x)

∫e^2xcos(5x)dx = (5/29)e^2xsin(5x) + (2/29)e^2xcos(5x) + C