Question

Let [a] denote the greatest integer which is less than or equal to a. Then the value of the integral ${\int }_{\frac{\pi }{2}}^{\frac{\pi }{2}}\left[sinxcosx\right]dx$ is

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

Answer 1

The correct option is D $-\frac{\pi }{2}$

${\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\left[\sin x\cos x\right]dx$

We know for,

$x\in (0,\frac{\pi }{2})0<\sin x\cos x<1\left[\sin x\cos x\right]=0x\in (-\frac{\pi }{2}0)-1<\sin x\cos x<0\left[\sin x\cos x\right]=-1{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\left[\sin x\cos x\right]dx={\int }_{-\frac{\pi }{2}}^0\left[\sin x\cos x\right]dx+{\int }_0^{\frac{\pi }{2}}\left[\sin x\cos x\right]dx={\int }_{-\frac{\pi }{2}}^0(-1)dx={-\left[x\right]}_{-\frac{\pi }{2}}^0=-\frac{\pi }{2}$

Hence the correct answer is $-\frac{\pi }{2}$