Answer
$\displaystyle \frac{e^{t}}{1+\alpha^{2}}[\sin(\alpha t-3)-\alpha \cos(\alpha t-3)]+C$
Work Step by Step
Assuming $\alpha \neq 0$,
$\displaystyle \int e^{t}\sin(\alpha t-3)dt=\quad \left[\begin{array}{ll}
u=\alpha t -3 & \\
du=\alpha dt & dt=\dfrac{du}{\alpha}
\end{array}\right]$
$\displaystyle \frac{1}{\alpha}\int e^{(u+3)/\alpha}\sin udu=\dfrac{1}{\alpha}e^{3/\alpha}\int e^{(1/\alpha)u}\sin udu$
Table of integrals:
$\color{blue}{98. \displaystyle \quad\int e^{au}\sin budu=\frac{e^{\mathrm{a}u}}{a^{2}+b^{2}} (a\sin bu-b \cos bu)+C }$
$a=1/\alpha,\quad b=1$
$\displaystyle \frac{1}{\alpha}e^{3/\alpha}\cdot\frac{e^{(1/\alpha)u}}{(1/\alpha)^{2}+1^{2}}(\frac{1}{\alpha}\sin u-\cos u)+C$
$=\displaystyle \frac{1}{\alpha}e^{3/\alpha}\cdot e^{(u/\alpha)}\frac{\alpha^{2}}{1+\alpha^{2}}(\frac{1}{\alpha}\sin u-\cos u)+C$
$=\displaystyle \frac{e^{(u+3)/\alpha}}{1+\alpha^{2}}(\sin u-\alpha\cos u)+C$
... bring back $t$...
$=\displaystyle \frac{e^{(\alpha t-3+3)/\alpha}}{1+\alpha^{2}}[\sin(\alpha t-3)-\alpha \cos(\alpha t-3)]+C$
$=\displaystyle \frac{e^{t}}{1+\alpha^{2}}[\sin(\alpha t-3)-\alpha \cos(\alpha t-3)]+C$
