int_{frac{-pi}{2}}^{frac{pi}{2}} sin^{2} x , | vedclass

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(C) Let $f(x) = \sin^{2} x$. Since $f(-x) = [\sin(-x)]^{2} = (-\sin x)^{2} = \sin^{2} x$,the function $f(x)$ is an even function. Using the property $

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$\frac{\pi}{2}$

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(C) Let $f(x) = \sin^{2} x$. Since $f(-x) = [\sin(-x)]^{2} = (-\sin x)^{2} = \sin^{2} x$,the function $f(x)$ is an even function. Using the property $\int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx$ for an even function: $\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \sin^{2} x \, dx = 2 \int_{0}^{\frac{\pi}{2}} \sin^{2} x \, dx$. Using the identity $\sin^{2} x = \frac{1 - \cos 2x}{2}$: $= 2 \int_{0}^{\frac{\pi}{2}} \frac{1 - \cos 2x}{2} \, dx = \int_{0}^{\frac{\pi}{2}} (1 - \cos 2x) \, dx$. Integrating term by term: $= [x - \frac{\sin 2x}{2}]_{0}^{\frac{\pi}{2}}$. Evaluating at the limits: $= (\frac{\pi}{2} - \frac{\sin \pi}{2}) - (0 - \frac{\sin 0}{2}) = (\frac{\pi}{2} - 0) - (0 - 0) = \frac{\pi}{2}$.

$\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \sin^{2} x \, dx =$

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