The value of the integral int_{-2}^{2} frac{s | vedclass

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(A) Let $I = \int_{-2}^{2} f(x) \, dx$,where $f(x) = \frac{\sin^2 x}{[\frac{x}{\pi}] + \frac{1}{2}}$.Since the interval is $[-2, 2]$ and $\pi \approx

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(A) Let $I = \int_{-2}^{2} f(x) \, dx$,where $f(x) = \frac{\sin^2 x}{[\frac{x}{\pi}] + \frac{1}{2}}$.Since the interval is $[-2, 2]$ and $\pi \approx 3.14$,the value of $\frac{x}{\pi}$ lies in the range $(-\frac{2}{\pi}, \frac{2}{\pi}) \approx (-0.63, 0.63)$.Thus,for $x \in (-2, 0)$,$[\frac{x}{\pi}] = -1$,and for $x \in [0, 2)$,$[\frac{x}{\pi}] = 0$.We split the integral: $I = \int_{-2}^{0} \frac{\sin^2 x}{-1 + 0.5} \, dx + \int_{0}^{2} \frac{\sin^2 x}{0 + 0.5} \, dx$.$I = \int_{-2}^{0} \frac{\sin^2 x}{-0.5} \, dx + \int_{0}^{2} \frac{\sin^2 x}{0.5} \, dx$.$I = -2 \int_{-2}^{0} \sin^2 x \, dx + 2 \int_{0}^{2} \sin^2 x \, dx$.Using the property $\int_{-a}^{0} f(x) \, dx = \int_{0}^{a} f(-x) \, dx$ and since $\sin^2(-x) = \sin^2 x$:$I = -2 \int_{0}^{2} \sin^2 x \, dx + 2 \int_{0}^{2} \sin^2 x \, dx = 0$.

The value of the integral $\int_{-2}^{2} \frac{\sin^2 x}{[\frac{x}{\pi}] + \frac{1}{2}} \, dx$ (where $[x]$ denotes the greatest integer less than or equal to $x$) is

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