Integral of e^(4x)cos(x-6)?

barcelonian

What's the integral of e^(4x)cos(x-6)?

Do we use integration by parts or what? It seems like it goes on forever.

anonymous

By integration by parts, we have...

∫ e^(4x)cos(x - 6) dx

u = e^(4x)

du/dx = 4e^(4x)

dv = cos(x - 6) dx

v = sin(x - 6)

You get...

∫ e^(4x)cos(x - 6) dx = e^(4x)sin(x - 6) - 4 ∫ e^(4x)sin(x - 6) dx

Repeat the same steps...

u = e^(4x)

du = 4e^(4x)

dv = sin(x - 6) dx

v = -cos(x - 6)

You get...

∫ e^(4x)cos(x - 6) dx = e^(4x)sin(x - 6) - 4(-e^(4x)cos(x - 6) + ∫ 4e^(4x)cos(x - 6) dx)

∫ e^(4x)cos(x - 6) dx = e^(4x)sin(x - 6) + 4e^(4x)cos(x - 6) - 16 ∫ e^(4x)cos(x - 6) dx

Bring -16 ∫ e^(4x)cos(x - 6) dx to the left to get...

17 ∫ e^(4x)cos(x - 6) dx = e^(4x)sin(x - 6) + 4e^(4x)cos(x - 6)

Divide both sides by 17 and include the constant to get...

∫ e^(4x)cos(x - 6) dx = (e^(4x)sin(x - 6) + 4e^(4x)cos(x - 6))/17 + c

I hope this helps!

captain-matticus-landpiratesinc

Integrate by parts:

int(e^(4x) * cos(x - 6) * dx) = ?

u = e^(4x)

du = 4 * e^(4x) * dx

dv = cos(x - 6) * dx

v = sin(x - 6)

int(e^(4x) * cos(x - 6) * dx) = uv - int(v * du)

int(e^(4x) * cos(x - 6) * dx) = e^(4x) * sin(x - 6) - int(4 * e^(4x) * sin(x - 6) * dx)

int(e^(4x) * cos(x - 6) * dx) = e^(4x) * sin(x - 6) - 4 * int(e^(4x) * sin(x - 6) * dx)

u = e^(4x)

du = 4 * e^(4x) * dx

dv = sin(x - 6) * dx

v = -cos(x - 6)

int(e^(4x) * cos(x - 6) * dx) = e^(4x) * sin(x - 6) - 4 * (uv - int(v * du))

int(e^(4x) * cos(x - 6) * dx) = e^(4x) * sin(x - 6) - 4 * uv + 4 * int(v * du)

int(e^(4x) * cos(x - 6) * dx) = e^(4x) * sin(x - 6) - 4 * e^(4x) * (-cos(x - 6)) + 4 * int(4 * e^(4x) * (-cos(x - 6)) * dx)

int(e^(4x) * cos(x - 6) * dx) = e^(4x) * sin(x - 6) + 4 * e^(4x) * cos(x - 6) - 16 * int(e^(4x) * cos(x - 6) * dx)

int(e^(4x) * cos(x - 6) * dx) + 16 * int(e^(4x) * cos(x - 6) * dx) = e^(4x) * sin(x - 6) + 4 * e^(4x) * cos(x - 6)

17 * int(e^(4x) * cos(x - 6) * dx) = e^(4x) * (sin(x - 6) + 4 * cos(x - 6))

int(e^(4x) * cos(x - 6) * dx) = (1/17) * e^(4x) * (sin(x - 6) + 4 * cos(x - 6))

Add the constant of integration:

int(e^(4x) * cos(x - 6) * dx) = (1/17) * e^(4x) * (sin(x - 6) + 4 * cos(x - 6)) + C

There you go.

hunter

e^(4 x) cos(6-x)