Evaluate int_{-frac{pi}{4}}^{frac{pi}{4}} sin | vedclass

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(A) We observe that $f(x) = \sin^{2} x$ is an even function because $f(-x) = \sin^{2}(-x) = (-\sin x)^{2} = \sin^{2} x = f(x)$.Therefore,by the proper

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$\frac{\pi-2}{4}$

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(A) We observe that $f(x) = \sin^{2} x$ is an even function because $f(-x) = \sin^{2}(-x) = (-\sin x)^{2} = \sin^{2} x = f(x)$.Therefore,by the property of definite integrals $\int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx$,we get:$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sin^{2} x \, dx = 2 \int_{0}^{\frac{\pi}{4}} \sin^{2} x \, dx$Using the identity $\sin^{2} x = \frac{1 - \cos 2x}{2}$,we have:$= 2 \int_{0}^{\frac{\pi}{4}} \frac{1 - \cos 2x}{2} \, dx = \int_{0}^{\frac{\pi}{4}} (1 - \cos 2x) \, dx$$= \left[ x - \frac{\sin 2x}{2} \right]_{0}^{\frac{\pi}{4}}$$= \left( \frac{\pi}{4} - \frac{\sin(\frac{\pi}{2})}{2} \right) - (0 - 0)$$= \frac{\pi}{4} - \frac{1}{2} = \frac{\pi - 2}{4}$

Evaluate $\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sin^{2} x \, dx$

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