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the value of the integral 2 2 2 x x 1 2 dx where x
Question:
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 Main - 2019
JEE Main
Updated On:
Aug 15, 2024
4
4 - sin 4
sin 4
0
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The Correct Option is
D
Solution and Explanation
$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 $
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