Question Number 103563 by mohammad17 last updated on 15/Jul/20

find laplase transform of the function  f(t)=sin^2 t  cos^3 t  ?

Answered by MAB last updated on 15/Jul/20

f(t)=(1−cos^2 t)cos^3 t  f(t)=cos^3 t−cos^5 t  f(t)=(1/4)cos(3t)+(3/4)cos(t)−((1/(16))cos(5t)  +(5/(16))cos(3t)+(5/8)cos(t))  f(t)=(1/8)cos(t)−(1/(16))cos(3t)−(1/(16))cos(5t)  L{f}(s)=(s/(8s^2 +8))−(s/(16s^2 +144))−(s/(16s^2 +400))

Answered by mathmax by abdo last updated on 16/Jul/20

f(t)=sin^2 t cos^3 t  ⇒f(t) =sin^2 t .cos^2 t cost  =((1/2)sin(2t))^2 cost =(1/4) sin^2 (2t)cost  =(1/4)(((1−cos(4t))/2))cost =(1/8)cost−(1/8)cost cos(4t)  =(1/8)cost−(1/(16)){ cos(5t)+cos(3t)} ⇒L(f(t))=(1/8)L(cost)−(1/(16))L(cos(5t))  −(1/(16))L(cos(3t))  but L(cos(wt))=∫_0 ^∞  cos(wx)e^(−tx) dt =Re(∫_0 ^∞ e^(iwx−tx) dt and  ∫_0 ^∞ e^((−t+iw)x) dx =[(1/(−t+iw))e^((−t+iw)x) ]_0 ^∞  =((−1)/(−t+iw)) =(1/(t−iw)) =((t+iw)/(t^2  +w^2 )) ⇒  L(cos(wt)) =(t/(t^2  +w^2 )) ⇒  L(f(t)) =(t/(8(t^2  +1)))−(1/6)×(t/(t^2  +25))−(1/(16))×(t/(t^2  +9))

Commented bymathmax by abdo last updated on 16/Jul/20

error of typo  L(f(t)) =(t/(8(t^2  +1)))−(t/(16(t^2  +25)))−(t/(16(t^2  +9)))