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

• A

  $\frac{\pi}{2}$
• B

  $\pi$
• C

  $-\pi$
• Correct answer
  D

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

Solution

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 $\theta=\pi$ $\begin{aligned} \text { Then, } & I=\frac{1}{2} \int_{\pi}^{\pi}\left[\frac{1}{2} \sin \theta\right] d \theta \\ &=\frac{1}{2}\left[\int_{\pi}^{0}(-1) d x+\int_{0}^{\pi}(0) d x\right] \\ &=\frac{1}{2}[-x]_{\pi}^{2}+0=-\frac{\pi}{2} \end{aligned}$

Hence, the correct answer is option: 'D'.