
What is the value of \[\int_{ - \dfrac{1}{2}}^{\dfrac{1}{2}} {\cos x\ln \dfrac{{1 + x}}{{1 - x}}dx} \]?
A. 0
B. 1
C. 2
D. \[\ln 3\]
Answer
162.9k+ views
Hint: The given integration is a definite integral. Then we will check whether the function is an odd function or an even function. Then apply definite integration property to solve it.
Formula Used:Property of the definite integral:
\[\int\limits_{ - a}^a {f\left( x \right)dx} = \left\{ {\begin{array}{*{20}{c}}0&{{\rm{if f(x) is an odd function}}}\\{2\int_0^a {f\left( x \right)dx} }&{{\rm{if f(x) is an even function}}}\end{array}} \right.\]
Logarithm Rule:
\[\ln {a^b} = b\ln a\]
Complete step by step solution:Given definite integral is \[\int_{ - \dfrac{1}{2}}^{\dfrac{1}{2}} {\cos x\ln \dfrac{{1 + x}}{{1 - x}}dx} \].
Assume that, \[f\left( x \right) = \cos x\ln \dfrac{{1 + x}}{{1 - x}}\]
To check whether the given function is an odd function or even function, we will put x = -x in \[f\left( x \right)\].
\[f\left( { - x} \right) = \cos \left( { - x} \right)\ln \dfrac{{1 + \left( { - x} \right)}}{{1 - \left( { - x} \right)}}\]
\[ \Rightarrow f\left( { - x} \right) = \cos \left( { - x} \right)\ln \dfrac{{1 - x}}{{1 + x}}\]
We know that \[\cos \left( { - x} \right) = \cos x\].
\[ \Rightarrow f\left( { - x} \right) = \cos x\ln {\left( {\dfrac{{1 + x}}{{1 - x}}} \right)^{ - 1}}\]
Now applying the formula \[\ln {a^b} = b\ln a\]
\[ \Rightarrow f\left( { - x} \right) = - \cos x\ln \dfrac{{1 + x}}{{1 - x}}\]
\[ \Rightarrow f\left( { - x} \right) = - f\left( x \right)\]
Thus \[f\left( x \right)\] is an odd function.
Now we will apply the definite integral property \[\int\limits_{ - a}^a {f\left( x \right)dx} = \left\{ {\begin{array}{*{20}{c}}0&{{\rm{if f(x) is an odd function}}}\\{2\int_0^a {f\left( x \right)dx} }&{{\rm{if f(x) is an even function}}}\end{array}} \right.\].
Since \[\cos x\ln \dfrac{{1 + x}}{{1 - x}}\] is a odd function, then
\[\int_{ - \dfrac{1}{2}}^{\dfrac{1}{2}} {\cos x\ln \dfrac{{1 + x}}{{1 - x}}dx} = 0\]
Option ‘A’ is correct
Note: Students often make mistake when they put the value of \[\cos \left( { - x} \right)\]. They use \[\cos \left( { - x} \right) = - \cos x\] which is an incorrect formula. Since cosine function is an even function thus \[\cos \left( { - x} \right) = \cos x\].
Formula Used:Property of the definite integral:
\[\int\limits_{ - a}^a {f\left( x \right)dx} = \left\{ {\begin{array}{*{20}{c}}0&{{\rm{if f(x) is an odd function}}}\\{2\int_0^a {f\left( x \right)dx} }&{{\rm{if f(x) is an even function}}}\end{array}} \right.\]
Logarithm Rule:
\[\ln {a^b} = b\ln a\]
Complete step by step solution:Given definite integral is \[\int_{ - \dfrac{1}{2}}^{\dfrac{1}{2}} {\cos x\ln \dfrac{{1 + x}}{{1 - x}}dx} \].
Assume that, \[f\left( x \right) = \cos x\ln \dfrac{{1 + x}}{{1 - x}}\]
To check whether the given function is an odd function or even function, we will put x = -x in \[f\left( x \right)\].
\[f\left( { - x} \right) = \cos \left( { - x} \right)\ln \dfrac{{1 + \left( { - x} \right)}}{{1 - \left( { - x} \right)}}\]
\[ \Rightarrow f\left( { - x} \right) = \cos \left( { - x} \right)\ln \dfrac{{1 - x}}{{1 + x}}\]
We know that \[\cos \left( { - x} \right) = \cos x\].
\[ \Rightarrow f\left( { - x} \right) = \cos x\ln {\left( {\dfrac{{1 + x}}{{1 - x}}} \right)^{ - 1}}\]
Now applying the formula \[\ln {a^b} = b\ln a\]
\[ \Rightarrow f\left( { - x} \right) = - \cos x\ln \dfrac{{1 + x}}{{1 - x}}\]
\[ \Rightarrow f\left( { - x} \right) = - f\left( x \right)\]
Thus \[f\left( x \right)\] is an odd function.
Now we will apply the definite integral property \[\int\limits_{ - a}^a {f\left( x \right)dx} = \left\{ {\begin{array}{*{20}{c}}0&{{\rm{if f(x) is an odd function}}}\\{2\int_0^a {f\left( x \right)dx} }&{{\rm{if f(x) is an even function}}}\end{array}} \right.\].
Since \[\cos x\ln \dfrac{{1 + x}}{{1 - x}}\] is a odd function, then
\[\int_{ - \dfrac{1}{2}}^{\dfrac{1}{2}} {\cos x\ln \dfrac{{1 + x}}{{1 - x}}dx} = 0\]
Option ‘A’ is correct
Note: Students often make mistake when they put the value of \[\cos \left( { - x} \right)\]. They use \[\cos \left( { - x} \right) = - \cos x\] which is an incorrect formula. Since cosine function is an even function thus \[\cos \left( { - x} \right) = \cos x\].
Recently Updated Pages
Fluid Pressure - Important Concepts and Tips for JEE

JEE Main 2023 (February 1st Shift 2) Physics Question Paper with Answer Key

Impulse Momentum Theorem Important Concepts and Tips for JEE

Graphical Methods of Vector Addition - Important Concepts for JEE

JEE Main 2022 (July 29th Shift 1) Chemistry Question Paper with Answer Key

JEE Main 2023 (February 1st Shift 1) Physics Question Paper with Answer Key

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

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

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

JoSAA JEE Main & Advanced 2025 Counselling: Registration Dates, Documents, Fees, Seat Allotment & Cut‑offs

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

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

NEET 2025 – Every New Update You Need to Know

Verb Forms Guide: V1, V2, V3, V4, V5 Explained

NEET Total Marks 2025

1 Billion in Rupees
