The value of int_{-pi/2}^{pi/2} frac{dx}{[x] | vedclass

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(C) Let $I = \int_{-\pi/2}^{\pi/2} \frac{dx}{[x] + [\sin x] + 4}$.We split the integral at $x=0$:$I = \int_{-\pi/2}^{0} \frac{dx}{[x] + [\sin x] + 4}

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$\frac{3}{20}(4\pi - 3)$

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(C) Let $I = \int_{-\pi/2}^{\pi/2} \frac{dx}{[x] + [\sin x] + 4}$.We split the integral at $x=0$:$I = \int_{-\pi/2}^{0} \frac{dx}{[x] + [\sin x] + 4} + \int_{0}^{\pi/2} \frac{dx}{[x] + [\sin x] + 4}$.For $x \in [-\pi/2, 0)$,$[\sin x] = -1$ (except at $x=0$). For $x \in [0, \pi/2]$,$[\sin x] = 0$.$I = \int_{-\pi/2}^{-1} \frac{dx}{[x] - 1 + 4} + \int_{-1}^{0} \frac{dx}{[x] - 1 + 4} + \int_{0}^{1} \frac{dx}{[x] + 0 + 4} + \int_{1}^{\pi/2} \frac{dx}{[x] + 0 + 4}$.$I = \int_{-\pi/2}^{-1} \frac{dx}{-2+3} + \int_{-1}^{0} \frac{dx}{-1+3} + \int_{0}^{1} \frac{dx}{0+4} + \int_{1}^{\pi/2} \frac{dx}{1+4}$.$I = \int_{-\pi/2}^{-1} 1 dx + \int_{-1}^{0} \frac{1}{2} dx + \int_{0}^{1} \frac{1}{4} dx + \int_{1}^{\pi/2} \frac{1}{5} dx$.$I = (-1 - (-\pi/2)) + \frac{1}{2}(0 - (-1)) + \frac{1}{4}(1 - 0) + \frac{1}{5}(\pi/2 - 1)$.$I = \frac{\pi}{2} - 1 + \frac{1}{2} + \frac{1}{4} + \frac{\pi}{10} - \frac{1}{5}$.$I = \pi(\frac{1}{2} + \frac{1}{10}) + (-1 + 0.5 + 0.25 - 0.2)$.$I = \frac{6\pi}{10} - 0.45 = \frac{3\pi}{5} - \frac{9}{20} = \frac{12\pi - 9}{20} = \frac{3(4\pi - 3)}{20}$.

The value of $\int_{-\pi/2}^{\pi/2} \frac{dx}{[x] + [\sin x] + 4}$,where $[t]$ denotes the greatest integer less than or equal to $t$,is

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