Given the following s-domain function:
$$ F(s) = \frac{2s}{(s+1)(s^2+4)} $$
determine f(t)
Begin by rewriting the function as: $$ F(s) = \Big( \frac{2}{s+1} \Big) \Big( \frac{s}{s^2+4} \Big) $$ Recall how we arrived at the definition the convolution integral in the introduction page: $$ F(s) = G(s)H(s) = \mathcal{L}[f(t)] $$ ...and also that: $$ \begin{align} f(t) &= \mathcal{L^{-1}}[G(s)H(s)] \\\\ &= g(t)*h(t) \qquad, (Expression\;1) \end{align} $$ For our example problem, let: $$ G(s) = \frac{2}{s+1} \quad, \;\; H(s) = \frac{s}{s^2+4} $$ ...and if we take the Laplace transform of both terms, we get: $$ g(t) = 2e^{-t} \quad, \;\; h(t) = \cos(2t) $$ Now we plug g(t) and h(t) into expression #1: $$ f(t) = 2e^{-t}*\cos(2t) $$ ...and by the definition of the convolution integral , we have: $$ \begin{align} f(t) &= \int_{0}^{t} g(\tau)h(t-\tau)\;d\tau \\\\ &= \int_{0}^{t} 2e^{-\tau} \cos(2[t-\tau]) \;d\tau \\\\ f(t) &= 2\int_{0}^{t}e^{-\tau} \cos(2t-2\tau)\;d\tau \qquad,(eqn\;2) \end{align} $$
To solve the integral in equation #2, we will use integration by parts: $$ Let: $$ $$ \begin{align} u &= \cos(2t-2\tau) \\ du&= 2\sin(2t-2\tau) \;d\tau \\ dv&= e^{-\tau}\\ v &= \int e^{-\tau} = -e^{-\tau} \end{align} $$ ...and by the definition of intergration by parts: $$ \int u\; dv = uv - \int v \; du $$ Therefore: $$ \int e^{-\tau}\cos(2t-2\tau)\;d\tau = -e^{-\tau}\cos(2t-2\tau)-\int -e^{-\tau}[2\sin(2t-2\tau)] \;d\tau $$ $$ \begin{align} \qquad \qquad &= -e^{-\tau}\cos(2t-2\tau) + 2\int e^{-\tau}\sin(2t-2\tau) \; d\tau \\\\ \end{align} $$ Once again we use intergration by parts for the integral term in the expression immediately above. Let: $$ \begin{align} u &= \sin(2t-2\tau) \\ du&= -2\cos(2t-2\tau) \; d\tau \\ dv&= e^{-\tau}\\ v &= \int e^{-\tau} = -e^{-\tau} \end{align} $$ Continuing from where we left off: $$ \begin{align} \int e^{-\tau}\cos(2t-2\tau)\;d\tau &= -e^{-\tau}\cos(2t-2\tau) + 2\Big[-e^{-\tau}\sin(2t-2\tau) \\ &-\int(-e^{-\tau})(-2\cos[2t-2\tau])\;d\tau \Big]\\\\\\ &= -e^{-\tau}\cos(2t-2\tau) + 2\Big[ -e^{-\tau}\sin(2t-2\tau) \Big] \\ &-2\int e^{-t}\cos(2t-2\tau)\;d\tau \\\\\\ \underline{\int e^{-\tau} \cos(2t-2\tau)\;d\tau} &= -e^{-\tau}\cos(2t-2\tau) - 2e^{-\tau}\sin(2t-2\tau)\\ &\underline{-4\int e^{-\tau} \cos(2t-2\tau) \;d\tau} \\\\\\ \end{align} $$ Notice the like terms underlined on both sides of the equation. Bring them both to the left side: $$ \begin{align} 5\int e^{-\tau}\cos(2t-2\tau)\;d\tau &= -e^{-\tau}\cos(2t-2\tau) - 2e^{-\tau}\sin(2t-2\tau)\\\\ \int e^{-\tau}\cos(2t-2\tau)\;d\tau &= \frac{-1}{5}e^{-\tau}\cos(2t-2\tau) - \frac{2}{5}e^{-\tau}\sin(2t-2\tau) \quad,\;(Eqn\;3) \end{align} $$ Now that we have solved the integral, substitute equation #3 into equation #2 and evaluate it:
$$ \begin{align} f(t) &= 2\int_{0}^{t}e^{-\tau} \cos(2t-2\tau)\;d\tau \\\\ &= 2\Big[ \frac{-1}{5}e^{-\tau}\cos(2t-2\tau) - \frac{2}{5}e^{-\tau}\sin(2t-2\tau) \Big] \Big|_0^t \\\\ &= \Big[ \frac{-2}{5}e^{-\tau}\cos(2t-2\tau) - \frac{4}{5}e^{-\tau}\sin(2t-2\tau) \Big]\Big|_0^t \end{align} $$ $$ \begin{align} &= \Big[ \frac{-2}{5}e^{-t}\cos(2t-2t) - \frac{4}{5}e^{-t}\sin(2t-2t) \Big] - \Big[ \frac{-2}{5}e^{-0}\cos(2t-0) - \frac{4}{5}e^{-0}\sin(2t-0) \Big] \\\\ &= \Big[ \frac{-2}{5}e^{-t}\cos(0) - \frac{4}{5}e^{-t}\sin(0) \Big] - \Big[ \frac{-2}{5}(1)\cos(2t) - \frac{4}{5}(1)\sin(2t) \Big] \\\\ \end{align} $$ $$ \begin{align} &= \Big[ \frac{-2}{5}e^{-t} - 0 \Big] - \Big[ \frac{-2}{5}\cos(2t)-\frac{4}{5}\sin(2t) \Big] \end{align} $$
$$ f(t) = \frac{-2}{5}e^{-t} + \frac{2}{5}\cos(2t)+\frac{4}{5}\sin(2t) $$