JEE MainMathematics

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+4 marks−1 if incorrectNumericalPrevious-year question

Definite Integral of Floor Functions of Trigonometric Expressions

Q: At x = π/2, csc x = 1, so [csc x] = 1. What about other points where csc x = 2?

At x = π/6 and 5π/6, csc x = 2, but individual points have zero measure (length 0) in Riemann integration, so only the open interval (π/6, 5π/6) where 1 < csc x < 2 matters, giving [csc x] = 1 almost everywhere.

Q: Why is [t] + [-t] = -1 and not 0?

For any non-integer t, write t = n + f where n ∈ ℤ and 0 < f < 1. Then [t] = n and [-t] = [-n - f] = -n - 1. Thus [t] + [-t] = -1.

Let [t] denotes the greatest integer ≤ t. Then 2/π integral from π / 6 to 5 π / 6 of (8[cosec x]-5[cot x]) d x is equal to

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Step-by-step solutionView

Correct answer

The value of the definite integral is 14.

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01Given

Integral $I = \frac{2}{π} \int_{π / 6}^{5 π / 6} (8 [\csc x] - 5 [\cot x]) d x$ , where $[\cdot]$ is the greatest integer function.

02Approach

Split the integral into two parts: $I_{1} = \frac{16}{π} \int_{π / 6}^{5 π / 6} [\csc x] d x$ and $I_{2} = - \frac{10}{π} \int_{π / 6}^{5 π / 6} [\cot x] d x$ . For $I_{1}$ , determine values of $[\csc x]$ on intervals where $\csc x$ takes integer values. For $I_{2}$ , use King's property $\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$ and $[t] + [- t] = - 1$ (for non-integers).

03Execute

Evaluate $I_{1} = \frac{16}{π} \int_{π / 6}^{5 π / 6} [\csc x] d x$ : On $(π / 6, 5 π / 6)$ , $\sin x \in (1 / 2, 1]$ , so $\csc x \in [1, 2)$ . Specifically, at $x = π / 2$ , $\csc x = 1$ , and for all $x \in (π / 6, 5 π / 6) ∖ {π / 2}$ , $1 < \csc x < 2$ , so $[\csc x] = 1$ . Thus, $\int_{π / 6}^{5 π / 6} [\csc x] d x = 1 \cdot (5 π / 6 - π / 6) = 2 π / 3$ . Therefore, $I_{1} = \frac{16}{π} \cdot \frac{2 π}{3} = \frac{32}{3}$ .

04Execute

Evaluate $J = \int_{π / 6}^{5 π / 6} [\cot x] d x$ using King's property ( $x \to π - x$ ): $J = \int_{π / 6}^{5 π / 6} [\cot (π - x)] d x = \int_{π / 6}^{5 π / 6} [- \cot x] d x$ . Adding the two expressions: $2 J = \int_{π / 6}^{5 π / 6} ([\cot x] + [- \cot x]) d x$ . Since $[t] + [- t] = - 1$ for all non-integer $t$ , and $\cot x$ is an integer only at finitely many points (measure zero): $2 J = \int_{π / 6}^{5 π / 6} (- 1) d x = - 1 \cdot (\frac{5 π}{6} - \frac{π}{6}) = - \frac{2 π}{3} ⟹ J = - \frac{π}{3}$ . Then $I_{2} = - \frac{10}{π} J = - \frac{10}{π} (- \frac{π}{3}) = \frac{10}{3}$ .

05Execute

Combine $I_{1}$ and $I_{2}$ to find $I = I_{1} + I_{2} = \frac{32}{3} + \frac{10}{3} = \frac{42}{3} = 14$ .

✓Verify

Check cotangent integral directly: $\cot x$ takes values in $[- \sqrt{3}, \sqrt{3}]$ . - $x \in (π / 6, π / 4)$ : $\cot x \in (1, \sqrt{3}] ⟹ [\cot x] = 1$ , length $π / 12$ , contribution $π / 12$ . - $x \in (π / 4, π / 2)$ : $\cot x \in (0, 1) ⟹ [\cot x] = 0$ , length $π / 4$ , contribution $0$ . - $x \in (π / 2, 3 π / 4)$ : $\cot x \in (- 1, 0) ⟹ [\cot x] = - 1$ , length $π / 4$ , contribution $- π / 4$ . - $x \in (3 π / 4, 5 π / 6)$ : $\cot x \in [- \sqrt{3}, - 1) ⟹ [\cot x] = - 2$ , length $π / 12$ , contribution $- π / 6$ . Sum $= π / 12 + 0 - π / 4 - π / 6 = - π / 3$ . Matches King's property result perfectly.

Hints that build this answer step by step

Which overall strategy best simplifies the integral $\int_{π / 6}^{5 π / 6} (8 [\csc x] - 5 [\cot x]) d x$ ?
Analyze the range of $\csc x$ directly on $(π / 6, 5 π / 6)$ and apply the reflection property $\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$ to the $[\cot x]$ term.
For $x \in (π / 6, 5 π / 6)$ , what is the value of $[\csc x]$ almost everywhere?
1
Using $\int_{π / 6}^{5 π / 6} [\cot x] d x = \int_{π / 6}^{5 π / 6} [\cot (π - x)] d x$ , what is the value of $\int_{π / 6}^{5 π / 6} [\cot x] d x$ ?
$- \frac{π}{3}$
Combining the evaluations, what is the final value of $\frac{2}{π} (8 \int_{π / 6}^{5 π / 6} [\csc x] d x - 5 \int_{π / 6}^{5 π / 6} [\cot x] d x)$ ?
14

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 8 September 2026

Quick checks

Students also ask

x = π / 2

\csc x = 1

, so

[\csc x] = 1

. What about other points where

\csc x = 2

At $x = π / 6$ and $5 π / 6$ , $\csc x = 2$ , but individual points have zero measure (length 0) in Riemann integration, so only the open interval $(π / 6, 5 π / 6)$ where $1 < \csc x < 2$ matters, giving $[\csc x] = 1$ almost everywhere.

Why is

[t] + [- t] = - 1

and not

0

For any non-integer $t$ , write $t = n + f$ where $n \in Z$ and $0 < f < 1$ . Then $[t] = n$ and $[- t] = [- n - f] = - n - 1$ . Thus $[t] + [- t] = - 1$ .

Answer

The value of the definite integral is 14.

Why each option works or fails

Step-by-step solution

Given: Integral $I = \frac{2}{π} \int_{π / 6}^{5 π / 6} (8 [\csc x] - 5 [\cot x]) d x$ , where $[\cdot]$ is the greatest integer function.
Approach: Split the integral into two parts: $I_{1} = \frac{16}{π} \int_{π / 6}^{5 π / 6} [\csc x] d x$ and $I_{2} = - \frac{10}{π} \int_{π / 6}^{5 π / 6} [\cot x] d x$ . For $I_{1}$ , determine values of $[\csc x]$ on intervals where $\csc x$ takes integer values. For $I_{2}$ , use King's property $\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$ and $[t] + [- t] = - 1$ (for non-integers).
Execute: Evaluate $I_{1} = \frac{16}{π} \int_{π / 6}^{5 π / 6} [\csc x] d x$ : On $(π / 6, 5 π / 6)$ , $\sin x \in (1 / 2, 1]$ , so $\csc x \in [1, 2)$ . Specifically, at $x = π / 2$ , $\csc x = 1$ , and for all $x \in (π / 6, 5 π / 6) ∖ {π / 2}$ , $1 < \csc x < 2$ , so $[\csc x] = 1$ . Thus, $\int_{π / 6}^{5 π / 6} [\csc x] d x = 1 \cdot (5 π / 6 - π / 6) = 2 π / 3$ . Therefore, $I_{1} = \frac{16}{π} \cdot \frac{2 π}{3} = \frac{32}{3}$ .
Execute: Evaluate $J = \int_{π / 6}^{5 π / 6} [\cot x] d x$ using King's property ( $x \to π - x$ ): $J = \int_{π / 6}^{5 π / 6} [\cot (π - x)] d x = \int_{π / 6}^{5 π / 6} [- \cot x] d x$ . Adding the two expressions: $2 J = \int_{π / 6}^{5 π / 6} ([\cot x] + [- \cot x]) d x$ . Since $[t] + [- t] = - 1$ for all non-integer $t$ , and $\cot x$ is an integer only at finitely many points (measure zero): $2 J = \int_{π / 6}^{5 π / 6} (- 1) d x = - 1 \cdot (\frac{5 π}{6} - \frac{π}{6}) = - \frac{2 π}{3} ⟹ J = - \frac{π}{3}$ . Then $I_{2} = - \frac{10}{π} J = - \frac{10}{π} (- \frac{π}{3}) = \frac{10}{3}$ .
Execute: Combine $I_{1}$ and $I_{2}$ to find $I = I_{1} + I_{2} = \frac{32}{3} + \frac{10}{3} = \frac{42}{3} = 14$ .
Verify: Check cotangent integral directly: $\cot x$ takes values in $[- \sqrt{3}, \sqrt{3}]$ . - $x \in (π / 6, π / 4)$ : $\cot x \in (1, \sqrt{3}] ⟹ [\cot x] = 1$ , length $π / 12$ , contribution $π / 12$ . - $x \in (π / 4, π / 2)$ : $\cot x \in (0, 1) ⟹ [\cot x] = 0$ , length $π / 4$ , contribution $0$ . - $x \in (π / 2, 3 π / 4)$ : $\cot x \in (- 1, 0) ⟹ [\cot x] = - 1$ , length $π / 4$ , contribution $- π / 4$ . - $x \in (3 π / 4, 5 π / 6)$ : $\cot x \in [- \sqrt{3}, - 1) ⟹ [\cot x] = - 2$ , length $π / 12$ , contribution $- π / 6$ . Sum $= π / 12 + 0 - π / 4 - π / 6 = - π / 3$ . Matches King's property result perfectly.

Definite Integral of Floor Functions of Trigonometric Expressions

What feels right?

The value of the definite integral is 14.

Why each option works or fails

Solution

01Given

02Approach

03Execute

04Execute

05Execute

✓Verify

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