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(B) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{12(3+[x])}{3+[\sin x]+[\cos x]} dx$. We split the integral based on the values of $[x]$,$[\si

Vedclass · Accepted Answer

$11\pi+2$

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(B) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{12(3+[x])}{3+[\sin x]+[\cos x]} dx$. We split the integral based on the values of $[x]$,$[\sin x]$,and $[\cos x]$ in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$. For $x \in [-\frac{\pi}{2}, -1)$,$[x] = -2$,$[\sin x] = -1$,$[\cos x] = 0$. So,the integrand is $\frac{12(3-2)}{3-1+0} = \frac{12}{2} = 6$. For $x \in [-1, 0)$,$[x] = -1$,$[\sin x] = -1$,$[\cos x] = 0$. So,the integrand is $\frac{12(3-1)}{3-1+0} = \frac{24}{2} = 12$. For $x \in [0, 1)$,$[x] = 0$,$[\sin x] = 0$,$[\cos x] = 0$. So,the integrand is $\frac{12(3+0)}{3+0+0} = \frac{36}{3} = 12$. For $x \in [1, \frac{\pi}{2}]$,$[x] = 1$,$[\sin x] = 0$,$[\cos x] = 0$. So,the integrand is $\frac{12(3+1)}{3+0+0} = \frac{48}{3} = 16$. Thus,$I = \int_{-\frac{\pi}{2}}^{-1} 6 dx + \int_{-1}^{0} 12 dx + \int_{0}^{1} 12 dx + \int_{1}^{\frac{\pi}{2}} 16 dx$. $I = 6(-1 - (-\frac{\pi}{2})) + 12(0 - (-1)) + 12(1 - 0) + 16(\frac{\pi}{2} - 1)$. $I = 6(-1 + \frac{\pi}{2}) + 12(1) + 12(1) + 16(\frac{\pi}{2} - 1)$. $I = -6 + 3\pi + 12 + 12 + 8\pi - 16$. $I = 11\pi + 2$.

Let $[\cdot]$ denote the greatest integer function. Then $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left( \frac{12(3+[x])}{3+[\sin x]+[\cos x]} \right) dx$ is equal to:

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