Question

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 1

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.