JEE Main · 2026 · 04 Apr 2026 · Shift 1 · Definite Integrals

Let

\int_{- 2}^{2} (∣ sin x ∣ + [x sin x]) d x = 2 (3 - cos 2) + β

, where

[\cdot]

is the greatest integer function. Then

β sin (\frac{β}{2})

equals.

Answer: (B)

Solution

Both

∣ sin x ∣

and

[x sin x]

are even, so the integral is

2 \int_{0}^{2} (∣ sin x ∣ + [x sin x]) d x

. On

[0, 2]

sin x \geq 0

gives

\int_{0}^{2} ∣ sin x ∣ d x = 1 - cos 2

, while

x sin x

increases from

0

to about

1.8

, so

[x sin x] = 0

until

x sin x = 1

x = α

and equals

1

beyond, giving

\int_{0}^{2} [x sin x] d x = 2 - α

. Thus the integral equals

2 (3 - cos 2) - 2 α

, and comparison gives

β = - 2 α

. Since

α sin α = 1

β sin \frac{β}{2} = (- 2 α) sin (- α) = 2 α sin α = 2

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