Question

I need help with this integral.

w= the integral from 0 to 5

24e^-6t cos(2t) dt.

i found the the integration in the integral table.

(e^ax/a^2 + b^2) (a cos bx + b sin bx)

im having trouble finishing the problem from here.

Answer 1

Write cos(x) as the real part of
exp(ix)

The integral of exp(ax)cos(bx) is then the real part of the integral of
exp[(a + ib)x]

The integral is:

1/(a+ib) exp[(a+ib)x] =

(a-ib)/(a^2 + b^2) exp(ax)*
[cos(bx) + i sin(bx)]

The real part of this is:

exp(ax)/(a^2 + b^2)
[a cos(bx) + b sin(bx)]