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Q. The value of the integral 22sin2x[xπ]+12dx (where [x] denotes the greatest integer less than 20Cr or equal to x) is :

JEE MainJEE Main 2019Integrals

Solution:

I=22sin2x[xπ]+12dx
I=20(sin2x[xπ]+12+sin2(x)[xπ]+12)dx
([xπ]+[xπ]=1asxnπ)
I=20(sin2x[xπ]+12+sin2x1[xπ]+12)dx=0