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Q. The value of the integral $\int \limits^{2}_{-2} \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 :

JEE MainJEE Main 2019Integrals

Solution:

$I = \int^{2}_{-2} \frac{\sin^{2}x}{\left[\frac{x}{\pi}\right] + \frac{1}{2}} dx $
$ I = \int^{2}_{0} \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^{2}_{0} \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 $