For any non-integer , write where and . Then and . Thus .

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Integral Calculus: JEE Main Mathematics Question with Solution

Q: At , , so . What about other points where ?

At and , , but individual points have zero measure (length 0) in Riemann integration, so only the open interval where matters, giving almost everywhere.

Let

[\mathrm{t}]

denotes the greatest integer

\leq \mathrm{t}

. Then

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

02approach

Split the integral into two parts: $I_1 = \frac{16}{\pi} \int_{\pi/6}^{5\pi/6} [\csc x] dx$ and $I_2 = -\frac{10}{\pi} \int_{\pi/6}^{5\pi/6} [\cot x] dx$ . For $I_1$ , determine values of $[\csc x]$ on intervals where $\csc x$ takes integer values. For $I_2$ , use King's property $\int_a^b f(x)dx = \int_a^b f(a+b-x)dx$ and $[t] + [-t] = -1$ (for non-integers).

03execute

Evaluate $I_1 = \frac{16}{\pi} \int_{\pi/6}^{5\pi/6} [\csc x] dx$ : On $(\pi/6, 5\pi/6)$ , $\sin x \in (1/2, 1]$ , so $\csc x \in [1, 2)$ . Specifically, at $x = \pi/2$ , $\csc x = 1$ , and for all $x \in (\pi/6, 5\pi/6) \setminus \{\pi/2\}$ , $1 < \csc x < 2$ , so $[\csc x] = 1$ . Thus, $\int_{\pi/6}^{5\pi/6} [\csc x] dx = 1 \cdot (5\pi/6 - \pi/6) = 2\pi/3$ . Therefore, $I_1 = \frac{16}{\pi} \cdot \frac{2\pi}{3} = \frac{32}{3}$ .

04execute

Evaluate $J = \int_{\pi/6}^{5\pi/6} [\cot x] dx$ using King's property ( $x \to \pi - x$ ): $J = \int_{\pi/6}^{5\pi/6} [\cot(\pi - x)] dx = \int_{\pi/6}^{5\pi/6} [-\cot x] dx$ . Adding the two expressions: $2J = \int_{\pi/6}^{5\pi/6} ([\cot x] + [-\cot x]) dx$ . Since $[t] + [-t] = -1$ for all non-integer $t$ , and $\cot x$ is an integer only at finitely many points (measure zero): $2J = \int_{\pi/6}^{5\pi/6} (-1) dx = -1 \cdot \left(\frac{5\pi}{6} - \frac{\pi}{6}\right) = -\frac{2\pi}{3} \implies J = -\frac{\pi}{3}$ . Then $I_2 = -\frac{10}{\pi} J = -\frac{10}{\pi} \left(-\frac{\pi}{3}\right) = \frac{10}{3}$ .

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 $[-\sqrt{3}, \sqrt{3}]$ . - $x \in (\pi/6, \pi/4)$ : $\cot x \in (1, \sqrt{3}] \implies [\cot x] = 1$ , length $\pi/12$ , contribution $\pi/12$ . - $x \in (\pi/4, \pi/2)$ : $\cot x \in (0, 1) \implies [\cot x] = 0$ , length $\pi/4$ , contribution $0$ . - $x \in (\pi/2, 3\pi/4)$ : $\cot x \in (-1, 0) \implies [\cot x] = -1$ , length $\pi/4$ , contribution $-\pi/4$ . - $x \in (3\pi/4, 5\pi/6)$ : $\cot x \in [-\sqrt{3}, -1) \implies [\cot x] = -2$ , length $\pi/12$ , contribution $-\pi/6$ . Sum $= \pi/12 + 0 - \pi/4 - \pi/6 = -\pi/3$ . Matches King's property result perfectly.

Hints that build this answer step by step

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

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 8 September 2026

Quick checks

Students also ask

x = \pi/2

\csc x = 1

, so

[\csc x] = 1

. What about other points where

\csc x = 2

At $x = \pi/6$ and $5\pi/6$ , $\csc x = 2$ , but individual points have zero measure (length 0) in Riemann integration, so only the open interval $(\pi/6, 5\pi/6)$ 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 $n \in \mathbb{Z}$ 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 $I = \frac{2}{\pi} \int_{\pi / 6}^{5 \pi / 6}(8[\csc x]-5[\cot x]) dx$ , where $[\cdot]$ is the greatest integer function.
approach: Split the integral into two parts: $I_1 = \frac{16}{\pi} \int_{\pi/6}^{5\pi/6} [\csc x] dx$ and $I_2 = -\frac{10}{\pi} \int_{\pi/6}^{5\pi/6} [\cot x] dx$ . For $I_1$ , determine values of $[\csc x]$ on intervals where $\csc x$ takes integer values. For $I_2$ , use King's property $\int_a^b f(x)dx = \int_a^b f(a+b-x)dx$ and $[t] + [-t] = -1$ (for non-integers).
execute: Evaluate $I_1 = \frac{16}{\pi} \int_{\pi/6}^{5\pi/6} [\csc x] dx$ : On $(\pi/6, 5\pi/6)$ , $\sin x \in (1/2, 1]$ , so $\csc x \in [1, 2)$ . Specifically, at $x = \pi/2$ , $\csc x = 1$ , and for all $x \in (\pi/6, 5\pi/6) \setminus \{\pi/2\}$ , $1 < \csc x < 2$ , so $[\csc x] = 1$ . Thus, $\int_{\pi/6}^{5\pi/6} [\csc x] dx = 1 \cdot (5\pi/6 - \pi/6) = 2\pi/3$ . Therefore, $I_1 = \frac{16}{\pi} \cdot \frac{2\pi}{3} = \frac{32}{3}$ .
execute: Evaluate $J = \int_{\pi/6}^{5\pi/6} [\cot x] dx$ using King's property ( $x \to \pi - x$ ): $J = \int_{\pi/6}^{5\pi/6} [\cot(\pi - x)] dx = \int_{\pi/6}^{5\pi/6} [-\cot x] dx$ . Adding the two expressions: $2J = \int_{\pi/6}^{5\pi/6} ([\cot x] + [-\cot x]) dx$ . Since $[t] + [-t] = -1$ for all non-integer $t$ , and $x \in (\pi/6, 5\pi/6) \setminus \{\pi/2\}$ is an integer only at finitely many points (measure zero): $2J = \int_{\pi/6}^{5\pi/6} (-1) dx = -1 \cdot \left(\frac{5\pi}{6} - \frac{\pi}{6}\right) = -\frac{2\pi}{3} \implies J = -\frac{\pi}{3}$ . Then $I_2 = -\frac{10}{\pi} J = -\frac{10}{\pi} \left(-\frac{\pi}{3}\right) = \frac{10}{3}$ .
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 $[-\sqrt{3}, \sqrt{3}]$ . - $x \in (\pi/6, \pi/4)$ : $\cot x \in (1, \sqrt{3}] \implies [\cot x] = 1$ , length $\pi/12$ , contribution $\pi/12$ . - $x \in (\pi/4, \pi/2)$ : $\cot x \in (0, 1) \implies [\cot x] = 0$ , length $\pi/4$ , contribution $0$ . - $x \in (\pi/2, 3\pi/4)$ : $\cot x \in (-1, 0) \implies [\cot x] = -1$ , length $\pi/4$ , contribution $-\pi/4$ . - $x \in (3\pi/4, 5\pi/6)$ : $\cot x \in [-\sqrt{3}, -1) \implies [\cot x] = -2$ , length $\pi/12$ , contribution $-\pi/6$ . Sum $= \pi/12 + 0 - \pi/4 - \pi/6 = -\pi/3$ . Matches King's property result perfectly.

Integral Calculus: JEE Main Mathematics Question with Solution

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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