JEE MAIN - Mathematics (2019 - 11th January Morning Slot - No. 15)
The value of the integral $$\int\limits_{ - 2}^2 {{{{{\sin }^2}x} \over { \left[ {{x \over \pi }} \right] + {1 \over 2}}}} \,dx$$ (where [x] denotes the greatest integer less than or equal to x) is
0
4
4$$-$$ sin 4
sin 4
Explanation
I $$=$$ $$\int\limits_{ - 2}^2 {{{{{\sin }^2}x} \over { \left[ {{x \over \pi }} \right] + {1 \over 2}}}} \,dx$$
$${\rm I} = \int\limits_0^2 {\left( {{{{{\sin }^2}x} \over {\left[ {{x \over \pi }} \right] + {1 \over 2}}} + {{{{\sin }^2}\left( { - x} \right)} \over {\left[ { - {x \over \pi }} \right] + {1 \over 2}}}} \right)dx} $$
$$\left( {\left[ {{x \over \pi }} \right] + \left[ { - {x \over \pi }} \right] = - 1\,\,} \right.$$ as $$\left. {\matrix{ \, \cr \, \cr } x \ne n\pi } \right)$$
$${\rm I} = \int\limits_0^2 {\left( {{{{{\sin }^2}x} \over {\left[ {{x \over \pi }} \right] + {1 \over 2}}} + {{{{\sin }^2}x} \over { - 1 - \left[ {{x \over \pi }} \right] + {1 \over 2}}}} \right)dx = 0} $$
$${\rm I} = \int\limits_0^2 {\left( {{{{{\sin }^2}x} \over {\left[ {{x \over \pi }} \right] + {1 \over 2}}} + {{{{\sin }^2}\left( { - x} \right)} \over {\left[ { - {x \over \pi }} \right] + {1 \over 2}}}} \right)dx} $$
$$\left( {\left[ {{x \over \pi }} \right] + \left[ { - {x \over \pi }} \right] = - 1\,\,} \right.$$ as $$\left. {\matrix{ \, \cr \, \cr } x \ne n\pi } \right)$$
$${\rm I} = \int\limits_0^2 {\left( {{{{{\sin }^2}x} \over {\left[ {{x \over \pi }} \right] + {1 \over 2}}} + {{{{\sin }^2}x} \over { - 1 - \left[ {{x \over \pi }} \right] + {1 \over 2}}}} \right)dx = 0} $$
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