Question

If $\left[a\right]$ denote the greatest integer which is less than or equal to $a$.

Then, the value of the integral ${\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\left[\sin x\cos x\right]dx$ is?

• A. $\frac{\pi }{2}$
• B. $\pi$
• C. $-\pi$
• D. $-\frac{\pi }{2}$

Answer 1

Answer: (D)

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

Let $I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\left[\sin x\cos x\right]dx$
$={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\left[\frac{1}{2}\sin 2x\right]dx$
Put $\theta =2x$ $\Rightarrow$ $d\theta 2dx$
Also when $x=-\pi /2$, then $\theta =-\pi$
when $x=\frac{\pi }{2}$, then $\theta =\pi$
Then, $I=\frac{1}{2}{\int }_{-\pi }^{\pi }\left[\frac{1}{2}\sin \theta \right]d\theta$
$=\frac{1}{2}\left[{\int }_{-\pi }^0(-1)dx+{\int }_0^{\pi }\left(0\right)dx\right]$
$=\frac{1}{2}{[-x]}_{-\pi }^0+0=-\frac{\pi }{2}$