JEE Main 2026 — Definite Integration Question with Solution

Let $I={\int }_{-2}^2\left(|\sin x|+[x\sin x]\right)dx=I_1+I_2$, where
$I_1={\int }_{-2}^2|\sin x|dx$ and $I_2={\int }_{-2}^2[x\sin x]dx$.

Evaluating $I_1$:
Since $|\sin x|$ is even, $I_1=2{\int }_0^2|\sin x|dx$.
For $x\in [0,2]$, $2<\pi$, so $\sin x\ge 0$ and $|\sin x|=\sin x$.
$I_1=2{\int }_0^2\sin xdx=2{\left[-\cos x\right]}_0^2=2(1-\cos 2)=2-2\cos 2$

Evaluating $I_2$:
Let $g(x)=x\sin x$. Then $g(-x)=(-x)\sin (-x)=x\sin x=g(x)$, so $g(x)$ is even, and hence $[x\sin x]$ is also even.
$I_2=2{\int }_0^2[x\sin x]dx$

Behaviour of $y=x\sin x$ on $[0,2]$:
At $x=0$, $y=0$. At $x=2$, $y=2\sin 2\approx 1.818$.
$y^{\prime }=\sin x+x\cos x>0$ on $[0,2]$, so $y$ is strictly increasing on this interval.

Therefore, $x\sin x$ takes each value in $[0,1.818]$ exactly once, and crosses $1$ at some unique $\alpha \in (0,2)$ with $\alpha \sin \alpha =1$.

For $x\in [0,\alpha )$: $0\le x\sin x<1\Rightarrow [x\sin x]=0$
For $x\in [\alpha ,2]$: $1\le x\sin x<2\Rightarrow [x\sin x]=1$

$I_2=2\left({\int }_0^{\alpha }0dx+{\int }_{\alpha }^{2}1dx\right)=2(2-\alpha )=4-2\alpha$

Combining:
$I=I_1+I_2=(2-2\cos 2)+(4-2\alpha )=6-2\cos 2-2\alpha$

Given $I=2(3-\cos 2)+\beta =6-2\cos 2+\beta$, so:
$6-2\cos 2-2\alpha =6-2\cos 2+\beta \Rightarrow \beta =-2\alpha$

Now:
$\beta \sin \left(\frac{\beta }{2}\right)=-2\alpha \sin \left(\frac{-2\alpha }{2}\right)=-2\alpha \sin (-\alpha )=2\alpha \sin \alpha$

Since $\alpha \sin \alpha =1$:
$\beta \sin \left(\frac{\beta }{2}\right)=2\cdot 1=2$

Hence, the correct option is $(2)2$.