Laplace transform and inverse laplace transform

Question

1. Find the Laplace transform of $$4e^2t-3\cos^2(2t)+2\cosh(3t).$$

My answer: $$\begin{align}\mathcal{L}(4e^2t-3\cos^2(2t)+2\cosh(3t))&=4\mathcal{L}(e^2t)-3\mathcal{L}(\cos^2(2t))+2\mathcal{L}(\cosh(3t))\\ &=\frac4 {s-2}-3\mathcal{L}(\cos^2(2t))+\frac{2s}{s^2-9}.\end{align}$$

But how can I find $\mathcal{L}(\cos^2(2t))$?

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2. Find inverse Laplace transform of:

i- $${6s-4}\over {s^2-4s+20}$$

ii- $${4s+12}\over {s^2+8s+16}$$

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Thanks.

Answer 1

1. Observing that $\cos^2(2t)=\frac{1+\cos(4t)}{2}$ we have $$\mathcal L\left\{\cos^2(2t)\right\}=\mathcal L\left\{\frac{1}{2}+\frac{1}{2}\cos(4t)\right\}=\frac{1}{2s}+\frac{1}{2}\frac{s}{s^2+16}$$ Thus the Laplace transform of $f(t)=4\mathrm e^{2t}-3\cos^2(2t)+2\cosh(3t)$ is $$ F(s)=\frac{4}{s-2}-\frac{3}{2s}-\frac{3}{2}\frac{s}{s^2+16}+2\frac{s}{s^2-9} $$
2. We can write $$G(s)=\frac{6s-4}{s^2-4s+20}=\frac{6(s-2)+8}{(s-2)^2+16}=6\frac{(s-2)}{(s-2)^2+16}+2\frac{4}{(s-2)^2+16}$$ and then $$ \mathcal L^{-1}\left\{G(s)\right\}=g(t)=6\mathrm e^{2t}\cos(4t)+2\mathrm e^{2t}\sin(4t)=2\mathrm e^{2t}\left[3\cos(4t)+\sin(4t)\right] $$
3. We can write $$H(s)=\frac{4s+12}{s^2+8s+16}=\frac{4(s+4)-4}{(s+4)^2}=\frac{4}{s+4}-\frac{4}{(s+4)^2}$$ and then $$ \mathcal L^{-1}\left\{H(s)\right\}=h(t)=4\mathrm e^{-4t}-4t\mathrm e^{-4t}=4\mathrm e^{-4t}(1-t) $$