Answer
$$0$$
Work Step by Step
Given $$\int_{-1}^{1}(\sin x)\left(\sin ^{2} x+1\right) d x$$
Since
$$f(x)=(\sin x)\left(\sin ^{2} x+1\right) $$
Then
\begin{align*}
f(-x)&=(\sin (-x))\left(\sin ^{2}(- x)+1\right) \\
&=-(\sin x)\left(\sin ^{2} x+1\right) \\
&=-f(x)
\end{align*}
which is an odd function, hence
$$\int_{-1}^{1}(\sin x)\left(\sin ^{2} x+1\right) d x=0$$
