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Let $f: R \rightarrow R$ and $g: R \rightarrow R$ be continuous functions. Then the value of the integral
$\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}[\mathrm{f}(x)+\mathrm{f}(-x)][\mathrm{g}(x)-\mathrm{g}(-x)] \mathrm{d} x$ is
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$\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}[\mathrm{f}(x)+\mathrm{f}(-x)][\mathrm{g}(x)-\mathrm{g}(-x)] \mathrm{d} x$ is
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Let $\mathrm{h}(x)=[\mathrm{f}(x)+\mathrm{f}(-x)][\mathrm{g}(x)-\mathrm{g}(-x)]$
$\begin{aligned}
\therefore \quad \mathrm{h}(-x) & =[\mathrm{f}(-x)+\mathrm{f}(x)][\mathrm{g}(-x)-\mathrm{g}(x)] \\
& =-[\mathrm{f}(x)+\mathrm{f}(-x)][\mathrm{g}(x)-\mathrm{g}(-x)] \\
& =-\mathrm{h}(x)
\end{aligned}$
$\therefore \quad \mathrm{h}(x)$ is an odd function.
$\therefore \quad \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \mathrm{~h}(x)=0$
$\begin{aligned}
\therefore \quad \mathrm{h}(-x) & =[\mathrm{f}(-x)+\mathrm{f}(x)][\mathrm{g}(-x)-\mathrm{g}(x)] \\
& =-[\mathrm{f}(x)+\mathrm{f}(-x)][\mathrm{g}(x)-\mathrm{g}(-x)] \\
& =-\mathrm{h}(x)
\end{aligned}$
$\therefore \quad \mathrm{h}(x)$ is an odd function.
$\therefore \quad \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \mathrm{~h}(x)=0$
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