Definite IntegrationmediumPYQ · JEE Main · 4 Apr 2026 · Shift 1 (Morning)Free

Definite Integration: Let Greatest Integer Function Equals (JEE Main 2026)

JEE Maths question with a full step-by-step solution.

Question
Let ∫−22(∣sin⁡x∣+[xsin⁡x])dx=2(3−cos⁡2)+β\displaystyle\int_{-2}^{2}\big(|\sin x|+[x\sin x]\big)dx=2(3-\cos2)+\beta, where [⋅][\cdot] is the greatest integer function. Then βsin⁡(β2)\beta\sin\left(\dfrac{\beta}{2}\right) equals
A11
B22correct
C44
D88
Solution
Step 1: The integrand is even, so
∫−22(∣sin⁡x∣+[xsin⁡x])dx=2∫02(∣sin⁡x∣+[xsin⁡x])dx=2(∫02sin⁡x dx+∫02[xsin⁡x] dx).\int_{-2}^{2}\big(|\sin x|+[x\sin x]\big)dx=2\int_0^2\big(|\sin x|+[x\sin x]\big)dx=2\left(\int_0^2\sin x\,dx+\int_0^2[x\sin x]\,dx\right).
 Step 2: On [0,2][0,2], xsin⁡xx\sin x ranges through [0,1][0,1]; let α\alpha be the root of xsin⁡x=1x\sin x=1. Then [xsin⁡x]=0[x\sin x]=0 on [0,α)[0,\alpha) and =1=1 on (α,2](\alpha,2]:
∫02[xsin⁡x] dx=∫α21 dx=2−α.\int_0^2[x\sin x]\,dx=\int_\alpha^2 1\,dx=2-\alpha.
 Step 3: So the integral =2([−cos⁡x]02+(2−α))=2((1−cos⁡2)+2−α)=2(3−cos⁡2)−2α.=2\big([-\cos x]_0^2+(2-\alpha)\big)=2\big((1-\cos2)+2-\alpha\big)=2(3-\cos2)-2\alpha. Comparing with 2(3−cos⁡2)+β2(3-\cos2)+\beta gives β=−2α\beta=-2\alpha, i.e. α=−β2\alpha=-\dfrac{\beta}{2}. Step 4: Since αsin⁡α=1\alpha\sin\alpha=1 and α=−β2\alpha=-\dfrac{\beta}{2}:
−β2sin⁡(−β2)=1 ⇒ β2sin⁡β2=1 ⇒ βsin⁡β2=2.-\frac{\beta}{2}\sin\left(-\frac{\beta}{2}\right)=1\ \Rightarrow\ \frac{\beta}{2}\sin\frac{\beta}{2}=1\ \Rightarrow\ \beta\sin\frac{\beta}{2}=2.
 Correct answer: (2)
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