use properties of the Laplace transform and the table of Laplace transforms to determine L[f] f(t)=int_0^t (t-w)cos(2w)dw

use properties of the Laplace transform and the table of Laplace transforms to determine L[f] f(t)=int_0^t (t-w)cos(2w)dw

Question
Laplace transform
asked 2021-03-02
use properties of the Laplace transform and the table of Laplace transforms to determine L[f] \(f(t)=\int_0^t (t-w)\cos(2w)dw\)

Answers (1)

2021-03-03
Step 1
We have given,
\(f(t)=\int_0^t (t-w)\cos(2w)dw\)
This can be written as,
\(f(t)=\int_0^t tcos(2w)dw-\int_0^t w\cos(2w)dw\)
We have to find,
\(L\left\{f(t)\right\}=L\left\{\int_0^t t\cos(2w)dw\right\}-L\left\{\int_0^t w\cos(2w)dw\right\}\)
Step 2
We know that,
\(L\left\{f(t)\right\}=F(s)\)
\(L\left\{\int_0^t f(u)du\right\}= \frac{1}{s}F(s)\)
So \(L\left\{\int_0^t t\cos(2w)dw\right\}=\frac{t}{s}(\frac{s}{s^2+4})\)
\(L\left\{\int_0^t t\cos(2w)dw\right\}=\frac{t}{s^2+4}\)
And
\(L\left\{\int_0^t w\cos(2w)dw\right\}=\frac{1}{s}\bigg(-\frac{-s^2+4}{(s^2+4)^2}\bigg)\)
On plugging these values in our equation , we get,
\(L\left\{f(t)\right\}=L\left\{\int_0^t t\cos(2w)dw\right\}-L\left\{\int_0^t w\cos(2w)dw\right\}\)
\(L\left\{f(t)\right\}=\bigg(\frac{t}{s^2+4}\bigg)+\frac{1}{s}\bigg(\frac{-s^2+4}{(s^2+4)^2}\bigg)\)
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