Let integral from -2 to 2 of (absolute value of sin x + greatest integer of x sin x) dx = 2(3 - cos 2) + beta, where [] is the greatest integer function. Then beta sin (beta/2) equals:

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q19 – JEE-MAIN 2026

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Home Maths IntegralsQuestion #207

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:

Clue 1 of 3: Integral Properties

Split the integral into two parts and use properties of even functions for symmetric limits, i.e., $\int_{- a}^{a} f (x) d x = 2 \int_{0}^{a} f (x) d x$ if $f (x)$ is even.

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● Video Walkthrough

Step 1: Split the integral and evaluate the first part✦ Active

The given integral is $I = \int_{- 2}^{2} (| \sin x | + [x \sin x]) d x$ . We can split it into two parts:

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

For the first part, $| \sin x |$ is an even function. Also, for $x \in [0, 2]$ , $\sin x \geq 0$ (since $2 radians \approx {114.6}^{\circ}$ , which is in the first or second quadrant). Thus, $| \sin x | = \sin x$ for $x \in [0, 2]$ . Using the property of even functions:

\int_{- 2}^{2} | \sin x | d x = 2 \int_{0}^{2} \sin x d x = 2 [- \cos x]_{0}^{2} = 2 (- \cos 2 - (- \cos 0)) = 2 (1 - \cos 2)

Step 2: Evaluate the second part of the integral○ Expand

Let $f (x) = x \sin x$ . Since $f (- x) = (- x) \sin (- x) = (- x) (- \sin x) = x \sin x = f (x)$ , $f (x)$ is an even function. Therefore, $[f (x)]$ is also an even function. Using the property of even functions:

\int_{- 2}^{2} [x \sin x] d x = 2 \int_{0}^{2} [x \sin x] d x

Now, let's analyze the range of $x \sin x$ for $x \in [0, 2]$ . At $x = 0$ , $x \sin x = 0$ . At $x = 2$ , $x \sin x = 2 \sin 2$ . Since $2 \sin 2 \approx 2 \times 0.909 = 1.818$ , the range of $x \sin x$ for $x \in [0, 2]$ is $[0, 2 \sin 2] \approx [0, 1.818]$ . This means that $[x \sin x]$ can only take integer values $0$ or $1$ in this interval.

Let $x_{0} \in (0, 2)$ be the unique value such that $x_{0} \sin x_{0} = 1$ . (Such $x_{0}$ exists because $x \sin x$ is continuous and increases from $0$ to $1.818$ in $[0, 2]$ ). Then:

- $[x \sin x] = 0$ for $x \in [0, x_{0})$ (where $0 \leq x \sin x < 1$ )

- $[x \sin x] = 1$ for $x \in [x_{0}, 2]$ (where $1 \leq x \sin x < 2 \sin 2 < 2$ )

So, the integral becomes:

2 \int_{0}^{2} [x \sin x] d x = 2 (\int_{0}^{x_{0}} 0 d x + \int_{x_{0}}^{2} 1 d x) = 2 ([x]_{x_{0}}^{2}) = 2 (2 - x_{0}) = 4 - 2 x_{0}

💡 Teacher's Secret Hint

Remember to check the behavior of $x \sin x$ over the entire interval $[0, 2]$ to correctly determine the values of the greatest integer function.

Step 3: Determine

β

and evaluate the final expression○ Expand

Substitute the evaluated parts back into the original equation:

2 (1 - \cos 2) + (4 - 2 x_{0}) = 2 (3 - \cos 2) + β

Expand and simplify:

2 - 2 \cos 2 + 4 - 2 x_{0} = 6 - 2 \cos 2 + β

6 - 2 \cos 2 - 2 x_{0} = 6 - 2 \cos 2 + β

This simplifies to:

β = - 2 x_{0}

We need to find the value of $β \sin (\frac{β}{2})$ . Substitute $β = - 2 x_{0}$ :

β \sin (\frac{β}{2}) = (- 2 x_{0}) \sin (\frac{- 2 x_{0}}{2}) = (- 2 x_{0}) \sin (- x_{0})

Since $\sin (- x) = - \sin x$ :

(- 2 x_{0}) (- \sin x_{0}) = 2 x_{0} \sin x_{0}

From Step 2, we defined $x_{0}$ such that $x_{0} \sin x_{0} = 1$ . Therefore:

2 x_{0} \sin x_{0} = 2 (1) = 2

💡 Teacher's Secret Hint

Notice how the unknown $x_{0}$ cancels out in the final expression, which is a common pattern in such problems.

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Maths Question 19 – JEE-MAIN 2026

Clue 1 of 3: Integral Properties

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