Answer
$$
\int_{-\pi / 2}^{\pi / 2} 3 \cos ^{3} x d x=4
$$
Work Step by Step
$$
\int_{-\pi / 2}^{\pi / 2} 3 \cos ^{3} x d x
$$
Since
\begin{align*}
\int_{-\pi / 2}^{\pi / 2} 3 \cos ^{3} x d x&=2\int_{0}^{\pi / 2} 3 \cos ^{3} x d x\\
&=6\int_{0}^{\pi / 2} \cos ^{2} x \cos x d x\\
&=6\int_{0}^{\pi / 2} (1-\sin^{2} x) \cos x d x\\
&=6 \left(\sin x-\frac{1}{3}\sin^{3} x\right)\bigg|_{0}^{\pi / 2}\\
&=6 \left(1-\frac{1}{3} \right)\\
&=4
\end{align*}
