What is Laplace transform of $frace^-tt sin3tsin2t$ [closed]

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What is Laplace of :

$$mathcalLleft(frace^-ttsin3tsin2tright)$$

I m trying to use first shifting property and not able to get correct answer.

calculus differential-equations laplace-transform laplacian

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edited Aug 2 at 18:15

user 108128

19k41544

asked Jul 29 at 17:07

Neetu yadav

112

closed as off-topic by amWhy, B. Mehta, Taroccoesbrocco, Adrian Keister, Mostafa Ayaz Jul 30 at 18:46

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, B. Mehta, Taroccoesbrocco, Adrian Keister, Mostafa Ayaz

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Isham what is answer?
  – Neetu yadav
  Jul 29 at 17:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I posted some hints to get the transform you want
  – Isham
  Jul 29 at 17:58
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

What is Laplace of :

$$mathcalLleft(frace^-ttsin3tsin2tright)$$

I m trying to use first shifting property and not able to get correct answer.

calculus differential-equations laplace-transform laplacian

share|cite|improve this question

edited Aug 2 at 18:15

user 108128

19k41544

asked Jul 29 at 17:07

Neetu yadav

112

closed as off-topic by amWhy, B. Mehta, Taroccoesbrocco, Adrian Keister, Mostafa Ayaz Jul 30 at 18:46

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, B. Mehta, Taroccoesbrocco, Adrian Keister, Mostafa Ayaz

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Isham what is answer?
  – Neetu yadav
  Jul 29 at 17:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I posted some hints to get the transform you want
  – Isham
  Jul 29 at 17:58
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

What is Laplace of :

$$mathcalLleft(frace^-ttsin3tsin2tright)$$

I m trying to use first shifting property and not able to get correct answer.

calculus differential-equations laplace-transform laplacian

share|cite|improve this question

edited Aug 2 at 18:15

user 108128

19k41544

asked Jul 29 at 17:07

Neetu yadav

112

What is Laplace of :

$$mathcalLleft(frace^-ttsin3tsin2tright)$$

I m trying to use first shifting property and not able to get correct answer.

calculus differential-equations laplace-transform laplacian

share|cite|improve this question

edited Aug 2 at 18:15

user 108128

19k41544

asked Jul 29 at 17:07

Neetu yadav

112

share|cite|improve this question

share|cite|improve this question

edited Aug 2 at 18:15

user 108128

19k41544

edited Aug 2 at 18:15

user 108128

19k41544

edited Aug 2 at 18:15

user 108128

19k41544

19k41544

asked Jul 29 at 17:07

Neetu yadav

112

asked Jul 29 at 17:07

Neetu yadav

112

asked Jul 29 at 17:07

Neetu yadav

112

112

closed as off-topic by amWhy, B. Mehta, Taroccoesbrocco, Adrian Keister, Mostafa Ayaz Jul 30 at 18:46

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, B. Mehta, Taroccoesbrocco, Adrian Keister, Mostafa Ayaz

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by amWhy, B. Mehta, Taroccoesbrocco, Adrian Keister, Mostafa Ayaz Jul 30 at 18:46

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, B. Mehta, Taroccoesbrocco, Adrian Keister, Mostafa Ayaz

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Isham what is answer?
  – Neetu yadav
  Jul 29 at 17:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I posted some hints to get the transform you want
  – Isham
  Jul 29 at 17:58
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Isham what is answer?
  – Neetu yadav
  Jul 29 at 17:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I posted some hints to get the transform you want
  – Isham
  Jul 29 at 17:58
  

  
  
  
  

@Isham what is answer?
– Neetu yadav
Jul 29 at 17:26

@Isham what is answer?
– Neetu yadav
Jul 29 at 17:26

I posted some hints to get the transform you want
– Isham
Jul 29 at 17:58

I posted some hints to get the transform you want
– Isham
Jul 29 at 17:58

add a comment | 

3 Answers
3

active

oldest

votes

up vote
0
down vote



accepted

$$frace^-tt sin(3t) sin(2t) = frace^-t2tcos(t) - frace^-t2tcos(5t)$$
Let $f(t) = e^-tcos(t)$ and $g(t) = e^-tcos(5t)$, we get
$$frace^-tt sin(3t) sin(2t) =frac12 (fracf(t)t -fracg(t)t)$$
Using the property that
$$L(fracf(t)t) = intlimits_s^infty F(u) du$$ we get
$$L(frace^-tt sin(3t) sin(2t)) = frac12[intlimits_s^infty F(u) du - intlimits_s^infty G(u) du]$$
where $F(s),G(s)$ are Laplace Transforms of $f(t),g(t)$.
Using $L(e^-atcos( wt) ) = fracs+a(s+a)^2 + w^2$, we can say
$$F(u) = fracs+1(s+1)^2 + 1 $$
$$G(u) = fracs+1(s+1)^2 + 25 $$
Computing the integrals:
$$intlimits_s^infty F(u) du = intlimits_s^infty fracu+1(u+1)^2 + 1 d u = frac12[lim_u rightarrow infty ln ((u+1)^2 + 1) - ln ((s+1)^2 + 1)]$$
$$intlimits_s^infty G(u) du = intlimits_s^infty fracu+1(u+1)^2 + 25 d u = frac12[lim_u rightarrow infty ln ((u+1)^2 + 25) - ln ((s+1)^2 + 25)]$$
Denote the limits as $A,B$, then:
$$L(frace^-tt sin(3t) sin(2t)) = frac14[ln ((s+1)^2 + 25)-ln ((s+1)^2 + 1) + A-B]$$
which is
$$L(frace^-tt sin(3t) sin(2t)) = frac14[ln frac(s+1)^2 + 25(s+1)^2 + 1 + lim_u rightarrow infty ln frac(u+1)^2 + 25(u+1)^2 + 1] = frac14ln frac(s+1)^2 + 25(s+1)^2 + 1$$

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answered Jul 29 at 17:37

Ahmad Bazzi

2,2531417

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0
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With integration of the formula $cal Lleft(e^ctright)=dfrac1s-c$ respect to $c$ we find
$$cal Lleft(dfrace^cttright)=lndfrac1s-c$$
then by $sinalpha=dfrace^ialpha-e^-ialpha2i$ we write
beginalign
cal Lleft(dfrace^-tsin3tsin2ttright)
&= -dfrac14cal Lleft(dfrace^(5i-1)t-e^(-i-1)t-e^(i-1)t+e^(-5i-1)ttright)\
&= -dfrac14cal Lleft(lndfrac1s-(5i-1)-lndfrac1s-(-i-1)-lndfrac1s-(i-1)+lndfrac1s-(-5i-1)right)\
&= colorblue-dfrac14lndfrac(s+1)^2+1(s+1)^2+25
endalign

share|cite|improve this answer

answered Jul 29 at 17:27

user 108128

19k41544

• 
  
  
  

  
  

  
  
  
  
  
  

  
  further can we put s=1 in final answer?
  – Neetu yadav
  Jul 29 at 17:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  For what purpose? it's a function of $s$.
  – user 108128
  Jul 29 at 17:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  answer will be -1/4 ln(5/29)?is it right?
  – Neetu yadav
  Jul 29 at 17:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I think it is a part of a problem.
  – user 108128
  Jul 29 at 17:35
  

  
  
  
  

add a comment | 

up vote
0
down vote

Hint

You can use
$$sin(a) sin(b)=frac 12( cos(a-b)-cos(a+b))$$
$$sin(3t) sin(2t)=frac 12( cos(t)-cos(5t))$$
Then use the formula
$$mathcalL left(frac f(t)tright )=int_s^infty F(u)du$$
I finally got this
$$F(s)=frac 14 ln left (frac s^2+2s+26s^2+2s+2 right)$$

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edited Jul 29 at 17:42

answered Jul 29 at 17:22

Isham

10.5k3829

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote



accepted

$$frace^-tt sin(3t) sin(2t) = frace^-t2tcos(t) - frace^-t2tcos(5t)$$
Let $f(t) = e^-tcos(t)$ and $g(t) = e^-tcos(5t)$, we get
$$frace^-tt sin(3t) sin(2t) =frac12 (fracf(t)t -fracg(t)t)$$
Using the property that
$$L(fracf(t)t) = intlimits_s^infty F(u) du$$ we get
$$L(frace^-tt sin(3t) sin(2t)) = frac12[intlimits_s^infty F(u) du - intlimits_s^infty G(u) du]$$
where $F(s),G(s)$ are Laplace Transforms of $f(t),g(t)$.
Using $L(e^-atcos( wt) ) = fracs+a(s+a)^2 + w^2$, we can say
$$F(u) = fracs+1(s+1)^2 + 1 $$
$$G(u) = fracs+1(s+1)^2 + 25 $$
Computing the integrals:
$$intlimits_s^infty F(u) du = intlimits_s^infty fracu+1(u+1)^2 + 1 d u = frac12[lim_u rightarrow infty ln ((u+1)^2 + 1) - ln ((s+1)^2 + 1)]$$
$$intlimits_s^infty G(u) du = intlimits_s^infty fracu+1(u+1)^2 + 25 d u = frac12[lim_u rightarrow infty ln ((u+1)^2 + 25) - ln ((s+1)^2 + 25)]$$
Denote the limits as $A,B$, then:
$$L(frace^-tt sin(3t) sin(2t)) = frac14[ln ((s+1)^2 + 25)-ln ((s+1)^2 + 1) + A-B]$$
which is
$$L(frace^-tt sin(3t) sin(2t)) = frac14[ln frac(s+1)^2 + 25(s+1)^2 + 1 + lim_u rightarrow infty ln frac(u+1)^2 + 25(u+1)^2 + 1] = frac14ln frac(s+1)^2 + 25(s+1)^2 + 1$$

share|cite|improve this answer

answered Jul 29 at 17:37

Ahmad Bazzi

2,2531417

add a comment | 

up vote
0
down vote



accepted

$$frace^-tt sin(3t) sin(2t) = frace^-t2tcos(t) - frace^-t2tcos(5t)$$
Let $f(t) = e^-tcos(t)$ and $g(t) = e^-tcos(5t)$, we get
$$frace^-tt sin(3t) sin(2t) =frac12 (fracf(t)t -fracg(t)t)$$
Using the property that
$$L(fracf(t)t) = intlimits_s^infty F(u) du$$ we get
$$L(frace^-tt sin(3t) sin(2t)) = frac12[intlimits_s^infty F(u) du - intlimits_s^infty G(u) du]$$
where $F(s),G(s)$ are Laplace Transforms of $f(t),g(t)$.
Using $L(e^-atcos( wt) ) = fracs+a(s+a)^2 + w^2$, we can say
$$F(u) = fracs+1(s+1)^2 + 1 $$
$$G(u) = fracs+1(s+1)^2 + 25 $$
Computing the integrals:
$$intlimits_s^infty F(u) du = intlimits_s^infty fracu+1(u+1)^2 + 1 d u = frac12[lim_u rightarrow infty ln ((u+1)^2 + 1) - ln ((s+1)^2 + 1)]$$
$$intlimits_s^infty G(u) du = intlimits_s^infty fracu+1(u+1)^2 + 25 d u = frac12[lim_u rightarrow infty ln ((u+1)^2 + 25) - ln ((s+1)^2 + 25)]$$
Denote the limits as $A,B$, then:
$$L(frace^-tt sin(3t) sin(2t)) = frac14[ln ((s+1)^2 + 25)-ln ((s+1)^2 + 1) + A-B]$$
which is
$$L(frace^-tt sin(3t) sin(2t)) = frac14[ln frac(s+1)^2 + 25(s+1)^2 + 1 + lim_u rightarrow infty ln frac(u+1)^2 + 25(u+1)^2 + 1] = frac14ln frac(s+1)^2 + 25(s+1)^2 + 1$$

share|cite|improve this answer

answered Jul 29 at 17:37

Ahmad Bazzi

2,2531417

add a comment | 

up vote
0
down vote



accepted

up vote
0
down vote



accepted

$$frace^-tt sin(3t) sin(2t) = frace^-t2tcos(t) - frace^-t2tcos(5t)$$
Let $f(t) = e^-tcos(t)$ and $g(t) = e^-tcos(5t)$, we get
$$frace^-tt sin(3t) sin(2t) =frac12 (fracf(t)t -fracg(t)t)$$
Using the property that
$$L(fracf(t)t) = intlimits_s^infty F(u) du$$ we get
$$L(frace^-tt sin(3t) sin(2t)) = frac12[intlimits_s^infty F(u) du - intlimits_s^infty G(u) du]$$
where $F(s),G(s)$ are Laplace Transforms of $f(t),g(t)$.
Using $L(e^-atcos( wt) ) = fracs+a(s+a)^2 + w^2$, we can say
$$F(u) = fracs+1(s+1)^2 + 1 $$
$$G(u) = fracs+1(s+1)^2 + 25 $$
Computing the integrals:
$$intlimits_s^infty F(u) du = intlimits_s^infty fracu+1(u+1)^2 + 1 d u = frac12[lim_u rightarrow infty ln ((u+1)^2 + 1) - ln ((s+1)^2 + 1)]$$
$$intlimits_s^infty G(u) du = intlimits_s^infty fracu+1(u+1)^2 + 25 d u = frac12[lim_u rightarrow infty ln ((u+1)^2 + 25) - ln ((s+1)^2 + 25)]$$
Denote the limits as $A,B$, then:
$$L(frace^-tt sin(3t) sin(2t)) = frac14[ln ((s+1)^2 + 25)-ln ((s+1)^2 + 1) + A-B]$$
which is
$$L(frace^-tt sin(3t) sin(2t)) = frac14[ln frac(s+1)^2 + 25(s+1)^2 + 1 + lim_u rightarrow infty ln frac(u+1)^2 + 25(u+1)^2 + 1] = frac14ln frac(s+1)^2 + 25(s+1)^2 + 1$$

share|cite|improve this answer

answered Jul 29 at 17:37

Ahmad Bazzi

2,2531417

$$frace^-tt sin(3t) sin(2t) = frace^-t2tcos(t) - frace^-t2tcos(5t)$$
Let $f(t) = e^-tcos(t)$ and $g(t) = e^-tcos(5t)$, we get
$$frace^-tt sin(3t) sin(2t) =frac12 (fracf(t)t -fracg(t)t)$$
Using the property that
$$L(fracf(t)t) = intlimits_s^infty F(u) du$$ we get
$$L(frace^-tt sin(3t) sin(2t)) = frac12[intlimits_s^infty F(u) du - intlimits_s^infty G(u) du]$$
where $F(s),G(s)$ are Laplace Transforms of $f(t),g(t)$.
Using $L(e^-atcos( wt) ) = fracs+a(s+a)^2 + w^2$, we can say
$$F(u) = fracs+1(s+1)^2 + 1 $$
$$G(u) = fracs+1(s+1)^2 + 25 $$
Computing the integrals:
$$intlimits_s^infty F(u) du = intlimits_s^infty fracu+1(u+1)^2 + 1 d u = frac12[lim_u rightarrow infty ln ((u+1)^2 + 1) - ln ((s+1)^2 + 1)]$$
$$intlimits_s^infty G(u) du = intlimits_s^infty fracu+1(u+1)^2 + 25 d u = frac12[lim_u rightarrow infty ln ((u+1)^2 + 25) - ln ((s+1)^2 + 25)]$$
Denote the limits as $A,B$, then:
$$L(frace^-tt sin(3t) sin(2t)) = frac14[ln ((s+1)^2 + 25)-ln ((s+1)^2 + 1) + A-B]$$
which is
$$L(frace^-tt sin(3t) sin(2t)) = frac14[ln frac(s+1)^2 + 25(s+1)^2 + 1 + lim_u rightarrow infty ln frac(u+1)^2 + 25(u+1)^2 + 1] = frac14ln frac(s+1)^2 + 25(s+1)^2 + 1$$

share|cite|improve this answer

answered Jul 29 at 17:37

Ahmad Bazzi

2,2531417

share|cite|improve this answer

share|cite|improve this answer

answered Jul 29 at 17:37

Ahmad Bazzi

2,2531417

answered Jul 29 at 17:37

Ahmad Bazzi

2,2531417

answered Jul 29 at 17:37

Ahmad Bazzi

2,2531417

2,2531417

add a comment | 

add a comment | 

up vote
0
down vote

With integration of the formula $cal Lleft(e^ctright)=dfrac1s-c$ respect to $c$ we find
$$cal Lleft(dfrace^cttright)=lndfrac1s-c$$
then by $sinalpha=dfrace^ialpha-e^-ialpha2i$ we write
beginalign
cal Lleft(dfrace^-tsin3tsin2ttright)
&= -dfrac14cal Lleft(dfrace^(5i-1)t-e^(-i-1)t-e^(i-1)t+e^(-5i-1)ttright)\
&= -dfrac14cal Lleft(lndfrac1s-(5i-1)-lndfrac1s-(-i-1)-lndfrac1s-(i-1)+lndfrac1s-(-5i-1)right)\
&= colorblue-dfrac14lndfrac(s+1)^2+1(s+1)^2+25
endalign

share|cite|improve this answer

answered Jul 29 at 17:27

user 108128

19k41544

• 
  
  
  

  
  

  
  
  
  
  
  

  
  further can we put s=1 in final answer?
  – Neetu yadav
  Jul 29 at 17:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  For what purpose? it's a function of $s$.
  – user 108128
  Jul 29 at 17:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  answer will be -1/4 ln(5/29)?is it right?
  – Neetu yadav
  Jul 29 at 17:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I think it is a part of a problem.
  – user 108128
  Jul 29 at 17:35
  

  
  
  
  

add a comment | 

up vote
0
down vote

With integration of the formula $cal Lleft(e^ctright)=dfrac1s-c$ respect to $c$ we find
$$cal Lleft(dfrace^cttright)=lndfrac1s-c$$
then by $sinalpha=dfrace^ialpha-e^-ialpha2i$ we write
beginalign
cal Lleft(dfrace^-tsin3tsin2ttright)
&= -dfrac14cal Lleft(dfrace^(5i-1)t-e^(-i-1)t-e^(i-1)t+e^(-5i-1)ttright)\
&= -dfrac14cal Lleft(lndfrac1s-(5i-1)-lndfrac1s-(-i-1)-lndfrac1s-(i-1)+lndfrac1s-(-5i-1)right)\
&= colorblue-dfrac14lndfrac(s+1)^2+1(s+1)^2+25
endalign

share|cite|improve this answer

answered Jul 29 at 17:27

user 108128

19k41544

• 
  
  
  

  
  

  
  
  
  
  
  

  
  further can we put s=1 in final answer?
  – Neetu yadav
  Jul 29 at 17:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  For what purpose? it's a function of $s$.
  – user 108128
  Jul 29 at 17:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  answer will be -1/4 ln(5/29)?is it right?
  – Neetu yadav
  Jul 29 at 17:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I think it is a part of a problem.
  – user 108128
  Jul 29 at 17:35
  

  
  
  
  

add a comment | 

up vote
0
down vote

up vote
0
down vote

With integration of the formula $cal Lleft(e^ctright)=dfrac1s-c$ respect to $c$ we find
$$cal Lleft(dfrace^cttright)=lndfrac1s-c$$
then by $sinalpha=dfrace^ialpha-e^-ialpha2i$ we write
beginalign
cal Lleft(dfrace^-tsin3tsin2ttright)
&= -dfrac14cal Lleft(dfrace^(5i-1)t-e^(-i-1)t-e^(i-1)t+e^(-5i-1)ttright)\
&= -dfrac14cal Lleft(lndfrac1s-(5i-1)-lndfrac1s-(-i-1)-lndfrac1s-(i-1)+lndfrac1s-(-5i-1)right)\
&= colorblue-dfrac14lndfrac(s+1)^2+1(s+1)^2+25
endalign

share|cite|improve this answer

answered Jul 29 at 17:27

user 108128

19k41544

With integration of the formula $cal Lleft(e^ctright)=dfrac1s-c$ respect to $c$ we find
$$cal Lleft(dfrace^cttright)=lndfrac1s-c$$
then by $sinalpha=dfrace^ialpha-e^-ialpha2i$ we write
beginalign
cal Lleft(dfrace^-tsin3tsin2ttright)
&= -dfrac14cal Lleft(dfrace^(5i-1)t-e^(-i-1)t-e^(i-1)t+e^(-5i-1)ttright)\
&= -dfrac14cal Lleft(lndfrac1s-(5i-1)-lndfrac1s-(-i-1)-lndfrac1s-(i-1)+lndfrac1s-(-5i-1)right)\
&= colorblue-dfrac14lndfrac(s+1)^2+1(s+1)^2+25
endalign

share|cite|improve this answer

answered Jul 29 at 17:27

user 108128

19k41544

share|cite|improve this answer

share|cite|improve this answer

answered Jul 29 at 17:27

user 108128

19k41544

answered Jul 29 at 17:27

user 108128

19k41544

answered Jul 29 at 17:27

user 108128

19k41544

19k41544

• 
  
  
  

  
  

  
  
  
  
  
  

  
  further can we put s=1 in final answer?
  – Neetu yadav
  Jul 29 at 17:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  For what purpose? it's a function of $s$.
  – user 108128
  Jul 29 at 17:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  answer will be -1/4 ln(5/29)?is it right?
  – Neetu yadav
  Jul 29 at 17:33
  

  
  
  
  


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  I think it is a part of a problem.
  – user 108128
  Jul 29 at 17:35
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  further can we put s=1 in final answer?
  – Neetu yadav
  Jul 29 at 17:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  For what purpose? it's a function of $s$.
  – user 108128
  Jul 29 at 17:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  answer will be -1/4 ln(5/29)?is it right?
  – Neetu yadav
  Jul 29 at 17:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I think it is a part of a problem.
  – user 108128
  Jul 29 at 17:35
  

  
  
  
  

further can we put s=1 in final answer?
– Neetu yadav
Jul 29 at 17:31

further can we put s=1 in final answer?
– Neetu yadav
Jul 29 at 17:31

For what purpose? it's a function of $s$.
– user 108128
Jul 29 at 17:32

For what purpose? it's a function of $s$.
– user 108128
Jul 29 at 17:32

answer will be -1/4 ln(5/29)?is it right?
– Neetu yadav
Jul 29 at 17:33

answer will be -1/4 ln(5/29)?is it right?
– Neetu yadav
Jul 29 at 17:33

I think it is a part of a problem.
– user 108128
Jul 29 at 17:35

I think it is a part of a problem.
– user 108128
Jul 29 at 17:35

add a comment | 

up vote
0
down vote

Hint

You can use
$$sin(a) sin(b)=frac 12( cos(a-b)-cos(a+b))$$
$$sin(3t) sin(2t)=frac 12( cos(t)-cos(5t))$$
Then use the formula
$$mathcalL left(frac f(t)tright )=int_s^infty F(u)du$$
I finally got this
$$F(s)=frac 14 ln left (frac s^2+2s+26s^2+2s+2 right)$$

share|cite|improve this answer

edited Jul 29 at 17:42

answered Jul 29 at 17:22

Isham

10.5k3829

add a comment | 

up vote
0
down vote

Hint

You can use
$$sin(a) sin(b)=frac 12( cos(a-b)-cos(a+b))$$
$$sin(3t) sin(2t)=frac 12( cos(t)-cos(5t))$$
Then use the formula
$$mathcalL left(frac f(t)tright )=int_s^infty F(u)du$$
I finally got this
$$F(s)=frac 14 ln left (frac s^2+2s+26s^2+2s+2 right)$$

share|cite|improve this answer

edited Jul 29 at 17:42

answered Jul 29 at 17:22

Isham

10.5k3829

add a comment | 

up vote
0
down vote

up vote
0
down vote

Hint

You can use
$$sin(a) sin(b)=frac 12( cos(a-b)-cos(a+b))$$
$$sin(3t) sin(2t)=frac 12( cos(t)-cos(5t))$$
Then use the formula
$$mathcalL left(frac f(t)tright )=int_s^infty F(u)du$$
I finally got this
$$F(s)=frac 14 ln left (frac s^2+2s+26s^2+2s+2 right)$$

share|cite|improve this answer

edited Jul 29 at 17:42

answered Jul 29 at 17:22

Isham

10.5k3829

Hint

You can use
$$sin(a) sin(b)=frac 12( cos(a-b)-cos(a+b))$$
$$sin(3t) sin(2t)=frac 12( cos(t)-cos(5t))$$
Then use the formula
$$mathcalL left(frac f(t)tright )=int_s^infty F(u)du$$
I finally got this
$$F(s)=frac 14 ln left (frac s^2+2s+26s^2+2s+2 right)$$

share|cite|improve this answer

edited Jul 29 at 17:42

answered Jul 29 at 17:22

Isham

10.5k3829

share|cite|improve this answer

share|cite|improve this answer

edited Jul 29 at 17:42

edited Jul 29 at 17:42

edited Jul 29 at 17:42

answered Jul 29 at 17:22

Isham

10.5k3829

answered Jul 29 at 17:22

Isham

10.5k3829

answered Jul 29 at 17:22

Isham

10.5k3829

10.5k3829

add a comment | 

add a comment |