If [a] denotes the greatest integer which is | vedclass

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(D) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} [\sin x \cos x] dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} [\frac{1}{2} \sin 2x] dx$. Substitute $\

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

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(D) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} [\sin x \cos x] dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} [\frac{1}{2} \sin 2x] dx$. Substitute $	heta = 2x$, so $d	heta = 2dx$ or $dx = \frac{1}{2} d	heta$. When $x = -\frac{\pi}{2}$, $	heta = -\pi$. When $x = \frac{\pi}{2}$, $	heta = \pi$. Thus, $I = \frac{1}{2} \int_{-\pi}^{\pi} [\frac{1}{2} \sin 	heta] d	heta$. Since $[\frac{1}{2} \sin 	heta]$ is an odd function, we check the intervals: For $	heta \in [-\pi, 0]$, $\sin 	heta \in [-1, 0]$, so $\frac{1}{2} \sin 	heta \in [-\frac{1}{2}, 0]$, hence $[\frac{1}{2} \sin 	heta] = -1$ for $	heta \in [-\pi, 0)$. For $	heta \in [0, \pi]$, $\sin 	heta \in [0, 1]$, so $\frac{1}{2} \sin 	heta \in [0, \frac{1}{2}]$, hence $[\frac{1}{2} \sin 	heta] = 0$ for $	heta \in [0, \pi]$. Therefore, $I = \frac{1}{2} [\int_{-\pi}^{0} (-1) d	heta + \int_{0}^{\pi} (0) d	heta] = \frac{1}{2} [-	heta]_{-\pi}^{0} = \frac{1}{2} [0 - (\pi)] = -\frac{\pi}{2}$.

If $[a]$ denotes the greatest integer which is less than or equal to $a$, then the value of the integral $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}[\sin x \cos x] dx$ is

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