Find the Laplace transform of this function $f(t)=\int _0^t(t-\tau)\cos(2\tau)d\tau$
My work:
$\mathcal{L}\{f(t)\}=\mathcal{L}\{\int_0^t(t-\tau)\cos(2\tau )d\tau\}$
Here I'm a little stuck, I think I need use
$\mathcal{L}\{\int_0 ^tf(t)dt\}=\dfrac{F(s)}{s}$
but I'm a little confused with that. Can someone help me?
You need to apply the convolution theorem. With the assumptions of the convolution theorem satisfied, if: $$f(t)=g(t){\ast}h(t)=\int_0^t{g(\tau)h(t-\tau)d\tau}=\int_0^t{g(t-\tau)h(\tau)d\tau}$$ where '$\ast$' implies convolution, then: $$F(s)=G(s)H(s)$$ In order to apply this theorem to your problem: Let $$f(t)=g(t){\ast}h(t)=\int_0^t{g(t-\tau)h(\tau)d\tau}$$ Conveniently choosing $g(t)=t$ and $h(t)=cos(2t)$ then $$g(t-\tau)=t-\tau$$ and $$h(\tau)=cos(2\tau)$$ Gives $$f(t)=g(t){\ast}h(t)=\int_0^t{(t-\tau)cos(2\tau)d\tau}$$ which is in the form of the problem to be solved.
Now, $$G(s)=\frac{1}{s^2}$$ and $$H(s)=\frac{s}{s^2+4}$$
Theferfore $$F(s)=G(s)H(s)=\frac{1}{s^2}\frac{s}{s^2+4}=\frac{1}{s(s^2+4)}$$