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If [a] denote the greatest integer which is less than or equal to a. Then, the value of the integral $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}[\sin x \cos x] d x$ is
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The correct answer is:
$-\frac{\pi}{2}$
Let $/=\int_{-\pi / 2}^{\pi / 2}[\sin x \cdot \cos x] d x=\int_{-\pi / 2}^{\pi / 2}\left[\frac{1}{2} \sin 2 x\right] d x$
Put $\theta=2 x \Rightarrow d \theta=2 d x$
Also, when $\quad x=-\pi / 2,$ then $\theta=-\pi$ when $\quad x=\frac{\pi}{2},$ then $\theta=\pi$
$\begin{aligned} \text { Then, } & I=\frac{1}{2} \int_{\pi}^{\pi}\left[\frac{1}{2} \sin \theta\right] d \theta \\ &=\frac{1}{2}\left[\int_{\pi}^{0}(-1) d x+\int_{0}^{\pi}(0) d x\right] \\ &=\frac{1}{2}[-x]_{\pi}^{2}+0=-\frac{\pi}{2} \end{aligned}$
Put $\theta=2 x \Rightarrow d \theta=2 d x$
Also, when $\quad x=-\pi / 2,$ then $\theta=-\pi$ when $\quad x=\frac{\pi}{2},$ then $\theta=\pi$
$\begin{aligned} \text { Then, } & I=\frac{1}{2} \int_{\pi}^{\pi}\left[\frac{1}{2} \sin \theta\right] d \theta \\ &=\frac{1}{2}\left[\int_{\pi}^{0}(-1) d x+\int_{0}^{\pi}(0) d x\right] \\ &=\frac{1}{2}[-x]_{\pi}^{2}+0=-\frac{\pi}{2} \end{aligned}$
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