Laplace transform of cos(bt)

recall that, cosbt = (e^ibt + e^-ibt)/2

Since L {e^(at)} = 1/(s - a),

L {cos(bt)} = L {(1/2)(e^(ibt) + e^(-ibt))}
................= (1/2) [L {(e^(ibt)} + L {e^(-ibt)}]
................= (1/2) [1/(s - bi) + 1/(s - (-bi))]
................= (1/2) * 2s/[(s - bi)(s + bi)]
................= s/(s^2 + b^2).

I hope this helps!