The value of the integral int_-2^2 (sin^2 x)/( [x/(π) ] + 1/2) dx (where, [x] denotes the greatest integer less than or equal to x) is

The value of the integral int_-2^2 (sin^2 x)/( [x/(π) ] +… — JEE Main 2019 Mathematics PYQ

JEE Main 2019 · 11 Jan · Shift I

The value of the integral

\int_{-2}^{2} \frac{\sin^2 x}{\left[\dfrac{x}{\pi}\right] + \dfrac{1}{2}} \, dx

(where,

[x]

denotes the greatest integer less than or equal to

x

) is

A.
$4 - \sin 4$
B.
$4$
C.
$\sin 4$
D.
$0$

Show answer and solution

Correct option: D

Solution

The integrand is f(x) = sin²x / ([x/π] + 1/2). Check if f(-x) = -f(x):

f(-x) = sin²(-x) / ([-x/π] + 1/2) = sin²x / ([-x/π] + 1/2).

For any non-integer t, [-t] = -[t] - 1, so [-x/π] = -[x/π] - 1, giving the denominator as -[x/π] - 1 + 1/2 = -([x/π] + 1/2).

Thus f(-x) = sin²x / (-([x/π] + 1/2)) = -f(x).

Since f is an odd function on the symmetric interval [-2, 2], the integral equals 0.

[JEE Main 2019 · 11 Jan · Shift I]

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