
What is the value of the definite integral \[\int\limits_0^1 {\dfrac{{\log x}}{{\sqrt {1 - {x^2}} }}} dx\]?
A. \[\dfrac{\pi }{2}\log 2\]
B. \[\pi \log 2\]
C. \[ - \dfrac{\pi }{2}\log 2\]
D. \[ - \pi \log 2\]
Answer
163.5k+ views
Hint: Here, a definite integral is given. First, substitute \[x = \sin \theta \] in the given integral and simplify it. Then, calculate the new limits of the integral. After that, apply the integration rule \[\int\limits_0^a {f\left( x \right)} dx = \int\limits_0^a {f\left( {a - x} \right)} dx\] and simplify the integral. Then, add both simplified integrals and solve them using the u-substitution method and trigonometric properties. In the end, apply the limits and get the required answer.
Formula Used:\[\int\limits_0^a {f\left( x \right)} dx = \int\limits_0^a {f\left( {a - x} \right)} dx\]
\[\log \left( a \right) + \log \left( b \right) = \log \left( {ab} \right)\]
\[\log \left( {\dfrac{a}{b}} \right) = \log \left( a \right) - \log \left( b \right)\]
\[\sin 2x = 2\sin x\cos x\]
\[{\sin ^2}x + {\cos ^2}x = 1\]
\[\int\limits_a^b {ndx} = \left[ {nx} \right]_a^b = n\left( {b - a} \right)\]
Complete step by step solution:The given definite integral is \[\int\limits_0^1 {\dfrac{{\log x}}{{\sqrt {1 - {x^2}} }}} dx\].
Let consider,
\[I = \int\limits_0^1 {\dfrac{{\log x}}{{\sqrt {1 - {x^2}} }}} dx\] \[.....\left( 1 \right)\]
Substitute \[x = \sin \theta \] in the above integral.
Differentiate the substituted equation.
\[dx = \cos \theta d\theta \]
Now calculate the now limits of the integral.
At \[x = 0\]:
\[\theta = {\sin ^{ - 1}}x\]
\[ \Rightarrow \theta = {\sin ^{ - 1}}0\]
\[ \Rightarrow \theta = 0\]
At \[x = 1\]:
\[\theta = {\sin ^{ - 1}}x\]
\[ \Rightarrow \theta = {\sin ^{ - 1}}1\]
\[ \Rightarrow \theta = \dfrac{\pi }{2}\]
Substitute the values in the equation \[\left( 1 \right)\].
\[I = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\log \sin \theta }}{{\sqrt {1 - {{\sin }^2}\theta } }}} \cos \theta d\theta \]
\[ \Rightarrow I = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\log \sin \theta }}{{\sqrt {{{\cos }^2}\theta } }}} \cos \theta d\theta \]
\[ \Rightarrow I = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\log \sin \theta }}{{\cos \theta }}} \cos \theta d\theta \]
\[ \Rightarrow I = \int\limits_0^{\dfrac{\pi }{2}} {\log \sin \theta } d\theta \] \[.....\left( 2 \right)\]
Now apply the integration rule \[\int\limits_0^a {f\left( x \right)} dx = \int\limits_0^a {f\left( {a - x} \right)} dx\].
\[ \Rightarrow I = \int\limits_0^{\dfrac{\pi }{2}} {\log \sin \left( {\dfrac{\pi }{2} - \theta } \right)d\theta } \]
\[ \Rightarrow I = \int\limits_0^{\dfrac{\pi }{2}} {\log \cos \theta d\theta } \] \[.....\left( 3 \right)\]
Add the equations \[\left( 2 \right)\] and \[\left( 3 \right)\].
\[ \Rightarrow I + I = \int\limits_0^{\dfrac{\pi }{2}} {\log \sin \theta d\theta } + \int\limits_0^{\dfrac{\pi }{2}} {\log \cos \theta d\theta } \]
Apply the sum rule of integration \[\int\limits_a^b {\left[ {f\left( x \right) + g\left( x \right)} \right]dx} = \int\limits_a^b {f\left( x \right)dx} + \int\limits_a^b {g\left( x \right)dx} \].
\[ \Rightarrow 2I = \int\limits_0^{\dfrac{\pi }{2}} {\left[ {\log \sin \theta + \log \cos \theta } \right]d\theta } \]
Apply the sum property of the logarithm \[\log \left( a \right) + \log \left( b \right) = \log \left( {ab} \right)\] .
\[ \Rightarrow 2I = \int\limits_0^{\dfrac{\pi }{2}} {\left[ {\log \sin \theta \cos \theta } \right]d\theta } \]
\[ \Rightarrow 2I = \int\limits_0^{\dfrac{\pi }{2}} {\log \left( {\dfrac{{2\sin \theta \cos \theta }}{2}} \right)d\theta } \]
\[ \Rightarrow 2I = \int\limits_0^{\dfrac{\pi }{2}} {\log \left( {\dfrac{{\sin 2\theta }}{2}} \right)d\theta } \]
Apply the quotient property of logarithm \[\log \left( {\dfrac{a}{b}} \right) = \log \left( a \right) - \log \left( b \right)\]
\[ \Rightarrow 2I = \int\limits_0^{\dfrac{\pi }{2}} {\log \left( {\dfrac{{\sin 2\theta }}{2}} \right)d\theta } \]
\[ \Rightarrow 2I = \int\limits_0^{\dfrac{\pi }{2}} {\left[ {\log \sin 2\theta - \log 2} \right]d\theta } \]
\[ \Rightarrow 2I = \int\limits_0^{\dfrac{\pi }{2}} {\log \sin 2\theta d\theta } - \int\limits_0^{\dfrac{\pi }{2}} {\log 2d\theta } \]
\[ \Rightarrow 2I = \int\limits_0^{\dfrac{\pi }{2}} {\log \sin 2\theta d\theta } - \log 2\int\limits_0^{\dfrac{\pi }{2}} {d\theta } \] \[.....\left( 4 \right)\]
Now substitute \[2\theta = u\] in the first integral.
Then, \[d\theta = \dfrac{{du}}{2}\]
The limits changed as follows:
As \[\theta \to 0\], then \[u \to 0\]
As \[\theta \to \dfrac{\pi }{2}\], then \[u \to \pi \]
Substitute the values in the equation \[\left( 4 \right)\].
\[2I = \dfrac{1}{2}\int\limits_0^\pi {\log \sin udu} - \log 2\int\limits_0^{\dfrac{\pi }{2}} {d\theta } \]
Apply the integration rule \[\int\limits_0^{na} {f\left( x \right)} dx = n\int\limits_0^a {f\left( x \right)} dx\].
\[2I = \int\limits_0^{\dfrac{\pi }{2}} {\log \sin udu} - \log 2\int\limits_0^{\dfrac{\pi }{2}} {d\theta } \]
From equation \[\left( 2 \right)\], we get \[I = \int\limits_0^{\dfrac{\pi }{2}} {\log \sin udu} \].
\[2I = I - \log 2\int\limits_0^{\dfrac{\pi }{2}} {d\theta } \]
\[ \Rightarrow I = - \log 2\left[ \theta \right]_0^{\dfrac{\pi }{2}}\]
\[ \Rightarrow I = - \log 2\left[ {\dfrac{\pi }{2} - 0} \right]\]
\[ \Rightarrow I = - \dfrac{\pi }{2}\log 2\]
Thus, \[\int\limits_0^1 {\dfrac{{\log x}}{{\sqrt {1 - {x^2}} }}} dx = - \dfrac{\pi }{2}\log 2\].
Option ‘C’ is correct
Note: Students get confused and try to solve the integral \[\int {\log \sin xdx} \] by using the formula \[\int {\log x = x\left( {\log x - 1} \right)} \] . Because of that, they get the wrong answer.
Formula Used:\[\int\limits_0^a {f\left( x \right)} dx = \int\limits_0^a {f\left( {a - x} \right)} dx\]
\[\log \left( a \right) + \log \left( b \right) = \log \left( {ab} \right)\]
\[\log \left( {\dfrac{a}{b}} \right) = \log \left( a \right) - \log \left( b \right)\]
\[\sin 2x = 2\sin x\cos x\]
\[{\sin ^2}x + {\cos ^2}x = 1\]
\[\int\limits_a^b {ndx} = \left[ {nx} \right]_a^b = n\left( {b - a} \right)\]
Complete step by step solution:The given definite integral is \[\int\limits_0^1 {\dfrac{{\log x}}{{\sqrt {1 - {x^2}} }}} dx\].
Let consider,
\[I = \int\limits_0^1 {\dfrac{{\log x}}{{\sqrt {1 - {x^2}} }}} dx\] \[.....\left( 1 \right)\]
Substitute \[x = \sin \theta \] in the above integral.
Differentiate the substituted equation.
\[dx = \cos \theta d\theta \]
Now calculate the now limits of the integral.
At \[x = 0\]:
\[\theta = {\sin ^{ - 1}}x\]
\[ \Rightarrow \theta = {\sin ^{ - 1}}0\]
\[ \Rightarrow \theta = 0\]
At \[x = 1\]:
\[\theta = {\sin ^{ - 1}}x\]
\[ \Rightarrow \theta = {\sin ^{ - 1}}1\]
\[ \Rightarrow \theta = \dfrac{\pi }{2}\]
Substitute the values in the equation \[\left( 1 \right)\].
\[I = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\log \sin \theta }}{{\sqrt {1 - {{\sin }^2}\theta } }}} \cos \theta d\theta \]
\[ \Rightarrow I = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\log \sin \theta }}{{\sqrt {{{\cos }^2}\theta } }}} \cos \theta d\theta \]
\[ \Rightarrow I = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\log \sin \theta }}{{\cos \theta }}} \cos \theta d\theta \]
\[ \Rightarrow I = \int\limits_0^{\dfrac{\pi }{2}} {\log \sin \theta } d\theta \] \[.....\left( 2 \right)\]
Now apply the integration rule \[\int\limits_0^a {f\left( x \right)} dx = \int\limits_0^a {f\left( {a - x} \right)} dx\].
\[ \Rightarrow I = \int\limits_0^{\dfrac{\pi }{2}} {\log \sin \left( {\dfrac{\pi }{2} - \theta } \right)d\theta } \]
\[ \Rightarrow I = \int\limits_0^{\dfrac{\pi }{2}} {\log \cos \theta d\theta } \] \[.....\left( 3 \right)\]
Add the equations \[\left( 2 \right)\] and \[\left( 3 \right)\].
\[ \Rightarrow I + I = \int\limits_0^{\dfrac{\pi }{2}} {\log \sin \theta d\theta } + \int\limits_0^{\dfrac{\pi }{2}} {\log \cos \theta d\theta } \]
Apply the sum rule of integration \[\int\limits_a^b {\left[ {f\left( x \right) + g\left( x \right)} \right]dx} = \int\limits_a^b {f\left( x \right)dx} + \int\limits_a^b {g\left( x \right)dx} \].
\[ \Rightarrow 2I = \int\limits_0^{\dfrac{\pi }{2}} {\left[ {\log \sin \theta + \log \cos \theta } \right]d\theta } \]
Apply the sum property of the logarithm \[\log \left( a \right) + \log \left( b \right) = \log \left( {ab} \right)\] .
\[ \Rightarrow 2I = \int\limits_0^{\dfrac{\pi }{2}} {\left[ {\log \sin \theta \cos \theta } \right]d\theta } \]
\[ \Rightarrow 2I = \int\limits_0^{\dfrac{\pi }{2}} {\log \left( {\dfrac{{2\sin \theta \cos \theta }}{2}} \right)d\theta } \]
\[ \Rightarrow 2I = \int\limits_0^{\dfrac{\pi }{2}} {\log \left( {\dfrac{{\sin 2\theta }}{2}} \right)d\theta } \]
Apply the quotient property of logarithm \[\log \left( {\dfrac{a}{b}} \right) = \log \left( a \right) - \log \left( b \right)\]
\[ \Rightarrow 2I = \int\limits_0^{\dfrac{\pi }{2}} {\log \left( {\dfrac{{\sin 2\theta }}{2}} \right)d\theta } \]
\[ \Rightarrow 2I = \int\limits_0^{\dfrac{\pi }{2}} {\left[ {\log \sin 2\theta - \log 2} \right]d\theta } \]
\[ \Rightarrow 2I = \int\limits_0^{\dfrac{\pi }{2}} {\log \sin 2\theta d\theta } - \int\limits_0^{\dfrac{\pi }{2}} {\log 2d\theta } \]
\[ \Rightarrow 2I = \int\limits_0^{\dfrac{\pi }{2}} {\log \sin 2\theta d\theta } - \log 2\int\limits_0^{\dfrac{\pi }{2}} {d\theta } \] \[.....\left( 4 \right)\]
Now substitute \[2\theta = u\] in the first integral.
Then, \[d\theta = \dfrac{{du}}{2}\]
The limits changed as follows:
As \[\theta \to 0\], then \[u \to 0\]
As \[\theta \to \dfrac{\pi }{2}\], then \[u \to \pi \]
Substitute the values in the equation \[\left( 4 \right)\].
\[2I = \dfrac{1}{2}\int\limits_0^\pi {\log \sin udu} - \log 2\int\limits_0^{\dfrac{\pi }{2}} {d\theta } \]
Apply the integration rule \[\int\limits_0^{na} {f\left( x \right)} dx = n\int\limits_0^a {f\left( x \right)} dx\].
\[2I = \int\limits_0^{\dfrac{\pi }{2}} {\log \sin udu} - \log 2\int\limits_0^{\dfrac{\pi }{2}} {d\theta } \]
From equation \[\left( 2 \right)\], we get \[I = \int\limits_0^{\dfrac{\pi }{2}} {\log \sin udu} \].
\[2I = I - \log 2\int\limits_0^{\dfrac{\pi }{2}} {d\theta } \]
\[ \Rightarrow I = - \log 2\left[ \theta \right]_0^{\dfrac{\pi }{2}}\]
\[ \Rightarrow I = - \log 2\left[ {\dfrac{\pi }{2} - 0} \right]\]
\[ \Rightarrow I = - \dfrac{\pi }{2}\log 2\]
Thus, \[\int\limits_0^1 {\dfrac{{\log x}}{{\sqrt {1 - {x^2}} }}} dx = - \dfrac{\pi }{2}\log 2\].
Option ‘C’ is correct
Note: Students get confused and try to solve the integral \[\int {\log \sin xdx} \] by using the formula \[\int {\log x = x\left( {\log x - 1} \right)} \] . Because of that, they get the wrong answer.
Recently Updated Pages
Geometry of Complex Numbers – Topics, Reception, Audience and Related Readings

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

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

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2025: Derivation of Equation of Trajectory in Physics

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

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

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

Degree of Dissociation and Its Formula With Solved Example for JEE

Instantaneous Velocity - Formula based Examples for JEE

JEE Advanced 2025 Notes

JEE Main Chemistry Question Paper with Answer Keys and Solutions
