Posted by Bobby Smith on Sunday, August 30, 2009 at 9:19pm.
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)]
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