Answer
$$\frac{8}{3}$$
Work Step by Step
We integrate as follows:
\begin{align*}
\int_{-\pi}^{\pi}\left(1-\cos ^{2} t\right)^{3 / 2} d t&=\int_{-\pi}^{\pi}\left(\sin ^{2} t\right)^{3 / 2} d t\\
&=\int_{-\pi}^{\pi}\left|\sin ^{3} t\right| d t\\
&=-\int_{-\pi}^{0} \sin ^{3} t d t+\int_{0}^{\pi} \sin ^{3} t d t\\
&=-\int_{-\pi}^{0}\left(1-\cos ^{2} t\right) \sin t d t+\int_{0}^{\pi}\left(1-\cos ^{2} t\right) \sin t d t\\
&=-\int_{-\pi}^{0} \sin t d t+\int_{-\pi}^{0} \cos ^{2} t \sin t d t+\int_{0}^{\pi} \sin t d t-\int_{0}^{\pi} \cos ^{2} t \sin t d t $
\\
&=\left[\cos t-\frac{\cos ^{3} t}{3}\right]_{-\pi}^{0}+\left[-\cos t+\frac{\cos ^{3} t}{3}\right]_{0}^{\pi}\\
&=\left(1-\frac{1}{3}+1-\frac{1}{3}\right)+\left(1-\frac{1}{3}+1-\frac{1}{3}\right)\\
&=\frac{8}{3}
\end{align*}
