#7,pg.311: Use the fact that $\cosh(bt)=\dfrac{e^{bt}+e^{-bt}}{2}$ to determine $\mathscr{L}\{\cosh(bt)\}(s)$.
Solution: We use the fact that $\mathscr{L}\{e^{at}\}(s)=\dfrac{1}{s-a}$ to see $$\begin{array}{ll} \mathscr{L}\{\cosh(bt)\}(s) &= \mathscr{L} \left\{ \dfrac{e^{bt}+e^{-bt}}{2} \right\}(s) \\ &=\dfrac{1}{2} \mathscr{L}\{e^{bt}\}(s) + \dfrac{1}{2} \mathscr{L}\{e^{-bt}\}(s) \\ &=\dfrac{1}{2} \dfrac{1}{s-b} + \dfrac{1}{2} \dfrac{1}{s-(-b)} \\ &=\dfrac{1}{2} \left[ \dfrac{1}{s-b} + \dfrac{1}{s+b} \right] \\ &= \dfrac{1}{2} \dfrac{2s}{s^2-b^2} \\ &= \dfrac{s}{s^2-b^2}. \end{array}$$
#13,pg.311: Use the fact that $\sin(bt)=\dfrac{e^{ibt}-e^{-ibt}}{2i}$ to compute $\mathscr{L}\{e^{at}\sin(bt)\}(s)$.
Solution: Use the fact that $\mathscr{L}\{e^{\xi t}\}(s)=\dfrac{1}{s-\xi}$ to see $$\begin{array}{ll} \mathscr{L}\{e^{at}\sin(bt)\}(s) &= \mathscr{L} \left\{e^{at} \dfrac{e^{ibt}-e^{-ibt}}{2i} \right\}(s) \\ &=\dfrac{1}{2i} \mathscr{L}\{e^{(a+ib)t}\}(s) - \dfrac{1}{2i} \mathscr{L}\{e^{(a-ib)t}\}(s) \\ &=\dfrac{1}{2i} \left[ \dfrac{1}{s-a-ib}-\dfrac{1}{s-a+ib} \right] \\ &= \dfrac{1}{2i} \dfrac{s-a+ib - (s-a-ib)}{(s-a)^2+b^2} \\ &=\dfrac{1}{2i} \dfrac{2ib}{(s-a)^2+b^2} \\ &= \dfrac{b}{(s-a)^2+b^2}. \end{array}$$