Question

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

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

Answer 1

Answer: (D)

$I={\int }_{-2}^2\frac{{\sin }^{2}x}{\left[\frac{x}{\pi }\right]+\frac{1}{2}}dx$
$I={\int }_0^2\left(\frac{{\sin }^{2}x}{\left[\frac{x}{\pi }\right]+\frac{1}{2}}+\frac{{\sin }^2\left(-x\right)}{\left[-\frac{x}{\pi }\right]+\frac{1}{2}}\right)dx$
$\left(\left[\frac{x}{\pi }\right]+\left[-\frac{x}{\pi }\right]=-1\text{as}x\ne n\pi \right)$
$I={\int }_0^2\left(\frac{{\sin }^{2}x}{\left[\frac{x}{\pi }\right]+\frac{1}{2}}+\frac{{\sin }^{2}x}{-1-\left[\frac{x}{\pi }\right]+\frac{1}{2}}\right)dx=0$