Question

# 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

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

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,
$$L\left\{f(t)\right\}=L\left\{\int_0^t t\cos(2w)dw\right\}-L\left\{\int_0^t w\cos(2w)dw\right\}$$
$$L\left\{f(t)\right\}=\bigg(\frac{t}{s^2+4}\bigg)+\frac{1}{s}\bigg(\frac{-s^2+4}{(s^2+4)^2}\bigg)$$