int_{-pi / 2}^{2 pi} sin ^{-1}(sin x) d x= | vedclass

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(B) Let $I = \int_{-\pi / 2}^{2 \pi} \sin ^{-1}(\sin x) d x$.We split the integral into intervals where $\sin^{-1}(\sin x)$ has a linear form:$I = \in

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$-\pi^2 / 8$

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(B) Let $I = \int_{-\pi / 2}^{2 \pi} \sin ^{-1}(\sin x) d x$.We split the integral into intervals where $\sin^{-1}(\sin x)$ has a linear form:$I = \int_{-\pi / 2}^{\pi / 2} x d x + \int_{\pi / 2}^{3 \pi / 2} (\pi - x) d x + \int_{3 \pi / 2}^{2 \pi} (x - 2 \pi) d x$Evaluating each integral:$\int_{-\pi / 2}^{\pi / 2} x d x = 0$ (since it is an odd function over a symmetric interval).$\int_{\pi / 2}^{3 \pi / 2} (\pi - x) d x = [\pi x - \frac{x^2}{2}]_{\pi / 2}^{3 \pi / 2} = (\frac{3 \pi^2}{2} - \frac{9 \pi^2}{8}) - (\frac{\pi^2}{2} - \frac{\pi^2}{8}) = \frac{3 \pi^2}{8} - \frac{3 \pi^2}{8} = 0$.$\int_{3 \pi / 2}^{2 \pi} (x - 2 \pi) d x = [\frac{x^2}{2} - 2 \pi x]_{3 \pi / 2}^{2 \pi} = (2 \pi^2 - 4 \pi^2) - (\frac{9 \pi^2}{8} - 3 \pi^2) = -2 \pi^2 - (-\frac{15 \pi^2}{8}) = -2 \pi^2 + \frac{15 \pi^2}{8} = -\frac{\pi^2}{8}$.Thus,$I = 0 + 0 - \frac{\pi^2}{8} = -\frac{\pi^2}{8}$.

$\int_{-\pi / 2}^{2 \pi} \sin ^{-1}(\sin x) d x=$

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