Let , where [.] is the greatest integer function. Then equals: [2026]

Question

Let $\int_{- 2}^{2} (| \sin x | + [x \sin x]) d x = 2 (3 - \cos 2) + β$ , where [.] is the greatest integer function. Then $β \sin (\frac{β}{2})$ equals: [2026]

Smart Achievers · Accepted Answer

(2)

We have, $\int_{- 2}^{2} (| \sin x | + [x \sin x]) d x$

$= \int_{- 2}^{2} | \sin x | d x + \int_{- 2}^{2} [x \sin x] d x$

$= 2 \int_{- 2}^{2} \sin x d x + 2 \int_{- 2}^{2} [x \sin x] d x$

$= 2 (- \cos x) |_{0}^{2} + 2 (\int_{0}^{α} 0 d x + \int_{α}^{2} d x)$ ( $∵ α is such that α \sin α = 1$ )

$= 2 (- \cos 2 + 1) + 2 (2 - α)$

$= 6 - 2 \cos 2 - 2 α$

On comparing, we get $β = - 2 α$

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

Solution

Q.
Let $\int_{- 2}^{2} (| \sin x | + [x \sin x]) d x = 2 (3 - \cos 2) + β$ , where [.] is the greatest integer function. Then $β \sin (\frac{β}{2})$ equals: [2026]

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Solution

Q. Let ∫–22(|sinx|+[x sinx])dx=2(3–cos2)+β, where [.] is the greatest integer function. Then βsin(β2) equals: [2026]

1 1

2 2

3 4

4 8

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Q.
Let $\int_{- 2}^{2} (| \sin x | + [x \sin x]) d x = 2 (3 - \cos 2) + β$ , where [.] is the greatest integer function. Then $β \sin (\frac{β}{2})$ equals: [2026]