Calculus 2 help?

Question

Evaluate the integral ( ^ means raise to )

1.) integration of: xe^ -x dx

2.) integration of: t sin 2t dt

Answer 1

Use integration by parts.

∫u dv = uv - ∫v du

∫xe^(-x) dx

u = x
dv = e^(-x) dx

du = dx
v = -e^(-x)

∫xe^(-x) dx = -xe^(-x) - ∫-e^(-x) dx = -xe^(-x) - e^(-x) + C
____________________

∫t sin(2t) dt

u = t
dv = sin(2t) dt

du = dt
v = -½cos(2t)

∫t sin(2t) dt = -½tcos(2t) - ∫-½cos(2t) dt
= -½tcos(2t) + ¼sin(2t) + C

Answer 2

integration by parts
int(v*du)=v*u-int(u*dv)

1. v=x, du=e^-x use these to find dv=1 and u=-e^-x
int(x*e^ -x) = x*-e^-x-int(-e^-x*1) = x*-e^-x-e^-x=-e^-x(x+1)

2. v=t, du=sin2t, dv=1, u=-1/2cos2t
int(t*sin2t) = t*-1/2cos2t-int(-1/2cos2t*1) = -t/2cos2t+1/4sin2t