Calculus
posted by Bobby Smith .
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.

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] =
(aib)/(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)]
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