Calculus 2 homework

Question

Evaluate the integral

8xcos(3x)dx

Answer 1

Perform integration by parts:

Let u = 8x meaning du = 8

Let dv = cos(3x) meaning v = (1/3)sin(3x)

Then, because ∫udv = uv - ∫vdu we get:

∫8xcos(3x) dx = (8x)[(1/3)sin(3x)] - ∫(8/3)sin(3x) dx

∫8xcos(3x) dx = (8/3)xsin(3x) + (8/9)cos(3x) + C