Question 4.1.4: Using Definition 4.1.1 Evaluate L {sin 2t}....
Using Definition 4.1.1
Evaluate \mathscr{L}\{\sin 2 t\}.
From Definition 4.1.1 and integration by parts we have
\begin{aligned} \mathscr{L}\{\sin 2 t\} \quad & \left.\int_{0}^{\infty} e^{-s t} \sin 2 t d t \quad \frac{-e^{-s t} \sin 2 t}{s}\right|_{0} ^{\infty}+\frac{2}{s} \int_{0}^{\infty} e^{-s t} \cos 2 t d t \\ & \frac{2}{s} \int_{0}^{\infty} e^{-s t} \cos 2 t d t, \quad s>0 \end{aligned}
\begin{array}{cc} \underset{t \rightarrow \infty}{\lim} e^{-s t} \cos 2 t=0, s>0 & \text { Laplace transform of } \sin 2 t \\ \downarrow & \downarrow \\ \frac{2}{s}\left[\left.\frac{-e^{-s t} \cos 2 t}{s}\right|_{0} ^{\infty}\right.- & \left.\frac{2}{s} \int_{0}^{\infty} e^{-s t} \sin 2 t d t\right] \end{array}
\frac{2}{s^{2}}-\frac{4}{s^{2}} \mathscr{L}\{\sin 2 t\}
At this point we have an equation with \mathscr{L}\{\sin 2 t\} on both sides of the equality. Solving for that quantity yields the result
\mathscr{L}\{\sin 2 t\} \quad \frac{2}{s^{2}+4}, \quad s>0.