laplace transformation chapter...sum

Question

Find L^-1  [sin2t.sin3t]   i need in simple steps solved . using 

laplace transformation. with formula which u applied please help me.....

Answer 1

cos(A+B)=cosAcosB-sinAsinB; cos(A-B)=cosAcosB+sinAsinB.

cos(A-B)-cos(A+B)=2sinAsinB, sinAsinB=½(cos(A-B)-cos(A+B)).

Let A=3t and B=2t, then A+B=5t and A-B=t.

sin(2t)sin(3t)=½(cos(t)-cos(5t)).

ℒ{cos(at)}=s/(s²+a²), so ℒ{cos(t)}=s/(s²+1); ℒ{cos(5t)}=s/(s²+25).

ℒ{sin(2t)sin(3t)}=½(s/(s²+1)-s/(s²+25))=

(s/2)(s²+25-s²-1)/(s⁴+26s²+25)=12s/(s⁴+26s²+25).

Laplace Transform of given expression is 12s/(s⁴+26s²+25).

ℒ^-1{12s/(s⁴+26s²+25)}=sin(2t)sin(3t).