Answer
$$T_3(x) =x-x^{2}+\frac{1}{3} x^{3}$$
Work Step by Step
Given $$f(x) =e^{-x}\sin x,\ \ a=0$$
Since
\begin{align*}
f(x) &=e^{-x}\sin x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ f(0) =0\\
f'(x) &=e^{-x}(\cos x-\sin x)\ \ \ \ \ \ \ f'(0 )=1\\
f''(x) &=-2e^{-x}\cos x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ f''(0 )=-2\\
f'''(x) &=-2e^{-x}(\cos x+\sin x)\ \ \ \ \ f'''(0 )=2
\end{align*}
Then
\begin{align*}
T_3(x)&= \sum_{n=0}^{3}\frac{f^{(n)}(0)}{n!}(x )^n\\
& =x-x^{2}+\frac{1}{3} x^{3}
\end{align*}
