Answer
converges
Work Step by Step
Given $$ \int_{8}^{\infty}\left(\sin ^{2} x\right) e^{-x} d x$$
Since for $x>0$
$$0 \leq\left(\sin ^{2} x\right) e^{-x} \leq e^{-x}$$
and
\begin{aligned}
\int_{8}^{\infty} e^{-x} d x &=\lim _{R \rightarrow \infty} \int_{8}^{R} e^{-x} d x \\
&=\lim _{R \rightarrow \infty}-\left.e^{-x}\right|_{8} ^{R} \\
&=\lim _{R \rightarrow \infty}\left(-e^{-R}+e^{-8}\right) \\
&=e^{-8}
\end{aligned}
Then by the comparison test, $ \int_{8}^{\infty}\left(\sin ^{2} x\right) e^{-x} d x$ also converges
