# Inverse Laplace Transform with squared irreducible quadratic in denominator using convolution theorem

$$F(s) = \dfrac{4}{(s^2+2s+5)^2}.$$

The answer I got is $\frac 1 5 (e^t - e^{-t}) \cos 2t - \frac 1 2 (e^t + e^{-t}) \sin 2t$.

I first applied the convolution theorem, then had to do integration by parts and the solution turned out to be too long is there any shorter solution for this problem, and is my answer correct??

Any help would be highly appreciated. Thanks.

First observe that $$F(s)=\frac{4}{\left[(s+1)^2+2^2\right]^2}$$ Let $\mathcal{L}^{-1}\left\{F(s)\right\}=f(t)$, then from the First Shifting Theorem we have
$$\mathcal{L}\left\{e^tf(t)\right\}=F(s-1)=\frac4{\left(s^2+2^2\right)^2}=\left(\frac2{s^2+2^2}\right)^2\tag{1}$$ Applying the Inverse Laplace Transform to the last equation we get, from the Convolution Theorem, \begin{align*} e^tf(t)&=\mathcal{L}^{-1}\left\{\frac2{s^2+2^2}\cdot\frac2{s^2+2^2}\right\}\\[4pt] &=\int_0^t\sin 2\tau\sin 2(t-\tau)d\tau\\[4pt]\tag{2} &=\int_0^t\frac12\left\{\cos\left[2\tau-2(t-\tau)\right]-\cos\left[2\tau+2(t-\tau)\right]\right\}d\tau\\[4pt] &=\int_0^t\frac12\left[\cos(4\tau-2t)-\cos(2t)\right]d\tau\\[4pt] &=\left.\frac18\sin(4\tau-2t)\right|_0^t-\frac12t\cos (2t)\\[4pt] &=\frac18\sin(4t-2t)-\frac18\sin(-2t)-\frac12t\cos (2t)\\[4pt] &=\frac14\sin(2t)-\frac12t\cos (2t) \end{align*} In the equation $(2)$ the identity $\color{red}{\sin x\sin y=\frac12\left[\cos(x-y)-\cos(x+y)\right]}$ have been used.
Then $$\boxed{\color{blue}{f(t)=\frac{e^{-t}(\sin 2t-2t\cos 2t)}{4}}}$$