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(D) Let $h(x) = [f(x)+f(-x)][g(x)-g(-x)]$. We evaluate $h(-x)$: $h(-x) = [f(-x)+f(x)][g(-x)-g(x)]$. This can be rewritten as: $h(-x) = [f(x)+f(-x)] \c

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(D) Let $h(x) = [f(x)+f(-x)][g(x)-g(-x)]$. We evaluate $h(-x)$: $h(-x) = [f(-x)+f(x)][g(-x)-g(x)]$. This can be rewritten as: $h(-x) = [f(x)+f(-x)] \cdot -[g(x)-g(-x)]$. $h(-x) = -[f(x)+f(-x)][g(x)-g(-x)] = -h(x)$. Since $h(-x) = -h(x)$,the function $h(x)$ is an odd function. According to the property of definite integrals,if $h(x)$ is an odd function,then $\int_{-a}^{a} h(x) \, dx = 0$. Therefore,$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} [f(x)+f(-x)][g(x)-g(-x)] \, dx = 0$.

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ and $g: \mathbb{R} \rightarrow \mathbb{R}$ be continuous functions. Then the value of the integral $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} [f(x)+f(-x)][g(x)-g(-x)] \, dx$ is

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