Find Laplace of given fn: f(t)=int_0^t sin(t-tau)costau d tau

EunoR 2020-11-27 Answered
Find Laplace of given fn:
f(t)=0tsin(tτ)cosτdτ
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Expert Answer

toroztatG
Answered 2020-11-28 Author has 98 answers
Step 1
Given that f(t)=0tsin(tτ)cosτdτ
The objective is to find the Laplace transform of f(t).
sin(tτ)=sintcosτcostsint
So,
sin(tτ)cosτ=sintcos2τcostsinτcosτ
=sint(1+cos2τ2)cost(sin2τ2)
Now, 
f(t)=sint20t[1+cos2τ]dτcost20tsin2τdτ
=sint2[t+sin2t2]cost2[cos2t2]0t
=tsint2+sin2tsint4cost2+costcos2t4
=tsint2cost2+14cos(t)
=tsint2cost4
Step 2
Now f(t)=tsint2cost4
So
L(f(t))=L(tsint2cost4)
=L(tsint2)+L(cost4)
=12L(tsint)14L(cost)
It is known that laplace transform of cost is s(s2+1)
Now let's find the L(tsint),
L[tsint]=d(ds)[L(sint)]
=d(ds)[1s2+1]
=2s(s2+1)2
Hence the laplace transform for f(t)=0tsin(tτ)cosτdτ is L(f(t))=s(s2+1)2s4(s2+1)
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