laplace of integral: Find L{int_{0,t} [e^{-3t} sin(4t) cos(2(t-tau)) d-tau}

[amarkant]

for this question do we 1st take the integral normally then do the laplace, i tried that and the calculation was very long. Is there an easier way to solve this?

 

[blamocur]

View attachment 36141
for this question do we 1st take the integral normally then do the laplace, i tried that and the calculation was very long. Is there an easier way to solve this?

I'd try "take the integral normally" first, then apply the Laplace transform. Have you tried this?

 

[amarkant]

I'd try "take the integral normally" first, then apply the Laplace transform. Have you tried this?

yes i tried that but the calculation was very long, was just wondering if there was an easier method

 

[nasi112]

View attachment 36141
for this question do we 1st take the integral normally then do the laplace, i tried that and the calculation was very long. Is there an easier way to solve this?

Solve the integral as blamocur suggested, you will get

\(\displaystyle \qquad \mathcal{L}\left\{\frac{1}{2}e^{-3t}\sin 4t\sin 2t\right\}\)

Use trigonometric identites

\(\displaystyle \qquad \sin 4t\sin 2t\ = \frac{1}{2}(\cos 2t - \cos 6t)\)

Now you have

\(\displaystyle \qquad \mathcal{L}\left\{\frac{1}{4}(e^{-3t}\cos 2t\ - e^{-3t}\cos 6t)\right\}\)

Look at the Laplace Transform table and you can answer this straightforward.

 

[blamocur]

yes i tried that but the calculation was very long, was just wondering if there was an easier method

I am not aware of easier methods, but feel free to post your efforts, and we might be able to help you move along.