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)]
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.
1 answer
1 answer