
$\int\limits_0^\pi {\left[ {\cot x} \right]} dx$where [.] denotes the greatest integer function, is equal to
$
{\text{A}}{\text{. }}\dfrac{\pi }{2} \\
{\text{B}}{\text{. 1}} \\
{\text{C}}{\text{. - 1}} \\
{\text{D}}{\text{. - }}\dfrac{\pi }{2} \\
$
Answer
152.1k+ views
Hint: To solve this question we have to understand property of greater integer function like: $
\left[ x \right] + \left[ { - x} \right] = 1,x \notin I \\
\left[ x \right] + \left[ { - x} \right] = 0,x \in I \\
$
Now we have to consider cot function on which x it gives integer value and on which other value and by doing this we can proceed further in the question.
Complete step-by-step answer:
In order to use the property,
$
\left[ x \right] + \left[ { - x} \right] = - 1,x \notin I \\
\left[ x \right] + \left[ { - x} \right] = 0,x \in I \\
$
We need to apply the below property of integers first-
$\int\limits_a^b {f\left( x \right)dx} = \int\limits_a^b {f\left( {b - a - x} \right)dx} $
This gives us:
$
\int\limits_0^\pi {\left[ {\cot x} \right]dx} = \int\limits_0^\pi {\left[ {\cot \left( {\pi - x} \right)} \right]dx} \\
\int\limits_0^\pi {\left[ {\cot \left( {\pi - x} \right)} \right]dx} = \int\limits_0^\pi {\left[ { - \cot x} \right]dx} \\
$
This shows:
$
I = \int\limits_0^\pi {\left[ {\cot x} \right]dx} \\
I = \int\limits_0^\pi {\left[ {\cot \left( {\pi - x} \right)} \right]dx} = \int\limits_0^\pi {\left[ { - \cot x} \right]dx} \\
$
Adding the two integrals-
$
2I = \int\limits_0^\pi {\left[ {\cot x} \right]dx} + \int\limits_0^\pi {\left[ { - \cot x} \right]dx} \\
2I = \int\limits_0^\pi {\left( {\left[ {\cot x} \right] + \left[ { - \cot x} \right]} \right)dx} \\
$
Now in the interval $0 - \pi $ cot(x) is not an integer so, from the property of greatest integer function we get,
$
2I = \int\limits_0^\pi {\left( {\left[ {\cot x} \right] + \left[ { - \cot x} \right]} \right)dx} \\
2I = \int_0^\pi { - 1.dx} \\
2I = - \left( {\pi - 0} \right) \\
2I = - \pi \\
I = \dfrac{{ - \pi }}{2} \\
$
The correct option is- D.
Note: Whenever we get this type of question the key concept of solving is we have to care more about negative values because on taking greater integers some students confuse. And we have to remember properties of definite integration like $\int\limits_a^b {f\left( x \right)dx} = \int\limits_a^b {f\left( {b - a - x} \right)dx} $.
\left[ x \right] + \left[ { - x} \right] = 1,x \notin I \\
\left[ x \right] + \left[ { - x} \right] = 0,x \in I \\
$
Now we have to consider cot function on which x it gives integer value and on which other value and by doing this we can proceed further in the question.
Complete step-by-step answer:
In order to use the property,
$
\left[ x \right] + \left[ { - x} \right] = - 1,x \notin I \\
\left[ x \right] + \left[ { - x} \right] = 0,x \in I \\
$
We need to apply the below property of integers first-
$\int\limits_a^b {f\left( x \right)dx} = \int\limits_a^b {f\left( {b - a - x} \right)dx} $
This gives us:
$
\int\limits_0^\pi {\left[ {\cot x} \right]dx} = \int\limits_0^\pi {\left[ {\cot \left( {\pi - x} \right)} \right]dx} \\
\int\limits_0^\pi {\left[ {\cot \left( {\pi - x} \right)} \right]dx} = \int\limits_0^\pi {\left[ { - \cot x} \right]dx} \\
$
This shows:
$
I = \int\limits_0^\pi {\left[ {\cot x} \right]dx} \\
I = \int\limits_0^\pi {\left[ {\cot \left( {\pi - x} \right)} \right]dx} = \int\limits_0^\pi {\left[ { - \cot x} \right]dx} \\
$
Adding the two integrals-
$
2I = \int\limits_0^\pi {\left[ {\cot x} \right]dx} + \int\limits_0^\pi {\left[ { - \cot x} \right]dx} \\
2I = \int\limits_0^\pi {\left( {\left[ {\cot x} \right] + \left[ { - \cot x} \right]} \right)dx} \\
$
Now in the interval $0 - \pi $ cot(x) is not an integer so, from the property of greatest integer function we get,
$
2I = \int\limits_0^\pi {\left( {\left[ {\cot x} \right] + \left[ { - \cot x} \right]} \right)dx} \\
2I = \int_0^\pi { - 1.dx} \\
2I = - \left( {\pi - 0} \right) \\
2I = - \pi \\
I = \dfrac{{ - \pi }}{2} \\
$
The correct option is- D.
Note: Whenever we get this type of question the key concept of solving is we have to care more about negative values because on taking greater integers some students confuse. And we have to remember properties of definite integration like $\int\limits_a^b {f\left( x \right)dx} = \int\limits_a^b {f\left( {b - a - x} \right)dx} $.
Recently Updated Pages
Difference Between Area and Volume

Difference Between Mutually Exclusive and Independent Events

JEE Main Participating Colleges 2024 - A Complete List of Top Colleges

JEE Main Maths Paper Pattern 2025 – Marking, Sections & Tips

Sign up for JEE Main 2025 Live Classes - Vedantu

JEE Main 2025 Helpline Numbers - Center Contact, Phone Number, Address

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Degree of Dissociation and Its Formula With Solved Example for JEE

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

Electrical Field of Charged Spherical Shell - JEE

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Displacement-Time Graph and Velocity-Time Graph for JEE

Collision - Important Concepts and Tips for JEE
