Answer
$$\int_{-\pi / 2}^{\pi / 2}\left(\sin ^{2} x+1\right) d x =\frac{3\pi}{2}$$
Work Step by Step
$$
\int_{-\pi / 2}^{\pi / 2}\left(\sin ^{2} x+1\right) d x
$$
Since
\begin{align*}
\int_{-\pi / 2}^{\pi / 2}\left(\sin ^{2} x+1\right) d x&=2\int_{0}^{\pi / 2}\left(\sin ^{2} x+1\right) d x\\
&=2\int_{0}^{\pi / 2}\left(\frac{3}{2} -\frac{1}{2}\cos 2 x\right) d x\\
&=2\left(\frac{3x}{2} -\frac{1}{4}\sin 2 x\right) \bigg|_{0}^{\pi / 2}\\
&=\frac{3\pi}{2}
\end{align*}
