Use Laplace transform to evaluate the integral int_0^infty te^{-2t} sin(2t)dt a) frac{1}{B} b) not defined c) frac{4s}{(s^2+4)^2} d) frac{4(s+2)}{(s^2+4s+8)^2}

emancipezN 2020-11-09 Answered
Use Laplace transform to evaluate the integral
0te2tsin(2t)dt
a) 1B
b) not defined
c) 4s(s2+4)2
d) 4(s+2)(s2+4s+8)2
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Expert Answer

Neelam Wainwright
Answered 2020-11-10 Author has 102 answers
Step 1
Here, the objective is to evaluate the given integral using Laplace transformation
0te2tsin(2t)dt
Step 2
According to the definition of Laplace transformation
F(s)=L(f(t))=0estf(t)ds
Compare the given integral with (1)
f(t)=tsin2t
To evaluate the given integral, find the Laplace transformation of f(t)
F(s)=L(f(t))
F(s)=L(tsin2t)
Since L(sin(at))=a(s2+a2)
Use L(tf(t))=dF(s)ds, Here f(t)=sin2tF(s)=a(s2+a2)
Therefore
F(s)=L(tsin2t)=dds(2(s2+22))
F(s)=(22s)(s2+4)2(where ddxxn=nxn1)
F(s)=4s(s2+4)2
Step 3
Hence
0testsin2tdt=4s(s2+4)2
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