How to calculate the Laplace Transform of the following
$f(t)=-t^3u_3(t)+\cos{(3t)}u_6(t)$ ...(1)
Solution:- The Laplace Transform of $\cos{(3t)}u_6(t)$ can be calculated using $\cos{(at+b)}=\frac{s\cos{(b)}-a\sin{(b)}}{s^2+a^2}$
Hence Laplace transform of the second term is $\bigg(\frac{s\cos{6}-3\sin{6}}{s^2+9}\bigg)e^{-6s}$
But how to compute the Laplace Transform of the first term?
If i write $f(t)=-(t-3+3)^3 u_3 (t)+cos{(3t-6+6)}u_6 (t)$.
$\bigg((t-3)^3+9(t-3)^2+27(t-3)+27)\bigg)=\bigg((t-3)^3+9t^2-27t+27\bigg)$
So the Laplace Transform of (1) is
$-\bigg(\frac{6}{s^4}+\frac{18}{s^3}-\frac{27}{s^2}+\frac{27}{s}\bigg)e^{-3s}+\bigg(\frac{s\cos{6}-3\sin{6}}{s^2+9}\bigg)e^{-6s}$
The Laplace transform of $\cos (3t) u_6(t)$ is correct. For the first term, you can take the Laplace transform of the expansion of $(3-t)^3$. Using the property of the Heaviside function (as we transform from $t$ to $s$): $$L\{f(t-a)u_a(t)\}=e^{-as}L\{f(t)\} \rightarrow (1)$$ $$a=3,f(t)=(3-t)^3 \Rightarrow L\{-t^3u_3(t)\}=e^{-3s}L\{(3-t)^3\}$$ $$\Rightarrow L\{-t^3u_3(t)\} = e^{-3s}L\{27-27t+9t^2-t^3\}$$ $$\Rightarrow L\{-t^3u_3(t)\} = e^{-3s}\left(\frac{27}{s}-\frac{27}{s^2}+\frac{18}{s^3}-\frac{6}{s^4}\right) \rightarrow (2)$$ Therefore we can conclude using $(2)$ and the laplace transform of the second term: $$L\{-t^3u_3(t)+\cos (3t) u_6(t)\}=3e^{-3s}\left(\frac{9}{s}-\frac{9}{s^2}+\frac{6}{s^3}-\frac{2}{s^4}\right)+e^{-6s}\left(\frac{s \cos 6 - 3 \sin 6}{s^2+9}\right)$$