2 math problems based on laplance transformation.

a) represent f(t) = sin2t, 2π < t < 4π and f(t) = 0 otherwise, in terms of unit step function and then find its laplance transform.
b) find the laplance transform of cos^2 t

[a]. f(t) = | sin2t, 2π < t < 4π
| 0, otherwise

f(t) = sin2t [u ( t - 2π ) - u ( t - 4π)]
L f(t) = L sin2t . u ( t - 2π ) - L sin2t . u ( t - 4π)
= e^(2πs) L sin2(t + 2π) - e^(-4πs) L sin2(t + 4π)
= e^(2πs) L sin2t - e^(-4πs) L sin(2t)
= e^(2πs) 2/(s² + 4) - e^(-4πs) 2/(s² + 4)
= [ e^(2πs) - e^(-4πs)] 2/(s² + 4)


[b]. cos2t = 2cos² t - 1
2cos² t = ½[cos2t + 1]

L (cos²t) = L [½ (cos2t + 1) ] = ½ [L (cos2t) + L (1)]
= ½[ s/(s² + 2²) + 1/s]
= ½[s/(s² + 4) + 1/s]

f(t)=sin2t(u(2pi-u(4pi))

L(f(t)=L(sin2t(u(2pi)-u(4pi)))

=(e(-2spi)-e(-4spi))L(sin2t)

=exp(-2pi*s)[1-exp(-2pi*s)]{4/(4+s^2}


L(cos^2t)

L(1+cos2t)/2

=L(1/2)+L(cos2t/2)

(1/2s)+(s/2s(s^2+4))

hope this helps