Question

The value of the integral $\stackrel{2}{\underset{-2}{\int }}\frac{{\sin }^{2}x}{\left[\frac{x}{\pi }\right]+\frac{1}{2}}dx$(where [x] denotes the greatest integer less than ${}^{20}Cr$ or equal to x) is:

• A. $4$
• B. $4-\sin 4$
• C. $\sin 4$
• D. $0$

Answer 1

Answer: (D)

$0$

$I=\stackrel{2}{\underset{-2}{\int }}$ $\frac{{\sin }^{2}x}{\left[\frac{x}{\pi }\right]+\frac{1}{2}}$
$I=\stackrel{2}{\underset{0}{\int }}(\frac{{\sin }^{2}x}{\left[\frac{x}{\pi }\right]+\frac{1}{2}}+\frac{{\sin }^2(-x)}{[-\frac{x}{\pi }]+\frac{1}{2}})dx$
$(\left[\frac{x}{\pi }\right]+[-\frac{x}{\pi }]=-1asx\ne n\pi )$
$I=\stackrel{2}{\underset{0}{\int }}(\frac{{\sin }^{2}x}{\left[\frac{x}{\pi }\right]+\frac{1}{2}}+\frac{{\sin }^{2}x}{-1-\left[\frac{x}{\pi }\right]+\frac{1}{2}})dx=0$
I=∫−22