Question

What is the value of the integral \[\int\limits_0^\pi {\log {{\sin }^2}xdx} \]?
A. \[2\pi {\log _e}\left( {\dfrac{1}{2}} \right)\]
B. \[\pi {\log _e}2 + c\]
C. \[\dfrac{\pi }{2}{\log _e}\left( {\dfrac{1}{2}} \right) + c\]
D. None of these

Answer 1

Hint: Here, a definite integral is given. First, simplify the integral by using the property \[\int\limits_0^{na} {f\left( x \right)} dx = n\int\limits_0^a {f\left( x \right)} dx\]. Rewrite \[{\sin ^2}x\] as \[{\left( {\sin x} \right)^2}\]. Then, apply the property of logarithm \[{\log _e}{\left( x \right)^n} = n{\log _e}\left( x \right)\] and simplify the term. 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, trigonometric and logarithmic properties. In the end, apply the limits and get the required answer.


Formula Used\[\int\limits_0^{na} {f\left( x \right)} dx = n\int\limits_0^a {f\left( x \right)} dx\]
\[\int\limits_0^a {f\left( x \right)} dx = \int\limits_0^a {f\left( {a - x} \right)} dx\]
\[\int\limits_a^b {\left[ {f\left( x \right) + g\left( x \right)} \right]dx} = \int\limits_a^b {f\left( x \right)dx} + \int\limits_a^b {g\left( x \right)dx} \]
\[\log \left( {\dfrac{a}{b}} \right) = \log \left( a \right) - \log \left( b \right)\]
\[\log \left( a \right) + \log \left( b \right) = \log \left( {ab} \right)\]



Complete step by step solution:The given definite integral us \[\int\limits_0^\pi {\log {{\sin }^2}xdx} \].
Let consider,
\[I = \int\limits_0^\pi {\log {{\sin }^2}xdx} \]
\[ \Rightarrow I = \int\limits_0^{2\dfrac{\pi }{2}} {\log {{\sin }^2}xdx} \]
Apply the integration rule \[\int\limits_0^{na} {f\left( x \right)} dx = n\int\limits_0^a {f\left( x \right)} dx\].
\[ \Rightarrow I = 2\int\limits_0^{\dfrac{\pi }{2}} {\log {{\sin }^2}xdx} \]
Rewrite \[{\sin ^2}x\] as \[{\left( {\sin x} \right)^2}\].
\[ \Rightarrow I = 2\int\limits_0^{\dfrac{\pi }{2}} {\log {{\left( {\sin x} \right)}^2}dx} \]
Apply the logarithmic property \[{\log _e}{\left( x \right)^n} = n{\log _e}\left( x \right)\].
\[ \Rightarrow I = 2\int\limits_0^{\dfrac{\pi }{2}} {2\log \sin xdx} \]
\[ \Rightarrow I = 4\int\limits_0^{\dfrac{\pi }{2}} {\log \sin xdx} \] \[.....\left( 1 \right)\]
Now apply the integration rule \[\int\limits_0^a {f\left( x \right)} dx = \int\limits_0^a {f\left( {a - x} \right)} dx\].
\[ \Rightarrow I = 4\int\limits_0^{\dfrac{\pi }{2}} {\log \sin \left( {\dfrac{\pi }{2} - x} \right)dx} \]
\[ \Rightarrow I = 4\int\limits_0^{\dfrac{\pi }{2}} {\log \cos xdx} \] \[.....\left( 2 \right)\]

Add the equations \[\left( 1 \right)\] and \[\left( 2 \right)\].
\[ \Rightarrow I + I = 4\int\limits_0^{\dfrac{\pi }{2}} {\log \sin xdx} + 4\int\limits_0^{\dfrac{\pi }{2}} {\log \cos xdx} \]
Apply the sum rule of integration \[\int\limits_a^b {\left[ {f\left( x \right) + g\left( x \right)} \right]dx} = \int\limits_a^b {f\left( x \right)dx} + \int\limits_a^b {g\left( x \right)dx} \].
\[ \Rightarrow 2I = 4\int\limits_0^{\dfrac{\pi }{2}} {\left[ {\log \sin x + \log \cos x} \right]dx} \]
Apply the sum property of the logarithm \[\log \left( a \right) + \log \left( b \right) = \log \left( {ab} \right)\] .
\[ \Rightarrow 2I = 4\int\limits_0^{\dfrac{\pi }{2}} {\left[ {\log \sin x\cos x} \right]dx} \]
\[ \Rightarrow 2I = 4\int\limits_0^{\dfrac{\pi }{2}} {\log \left( {\dfrac{{2\sin x\cos x}}{2}} \right)dx} \]
\[ \Rightarrow 2I = 4\int\limits_0^{\dfrac{\pi }{2}} {\log \left( {\dfrac{{\sin 2x}}{2}} \right)dx} \]
Apply the quotient property of logarithm \[\log \left( {\dfrac{a}{b}} \right) = \log \left( a \right) - \log \left( b \right)\]
\[ \Rightarrow 2I = 4\int\limits_0^{\dfrac{\pi }{2}} {\log \left( {\dfrac{{\sin 2x}}{2}} \right)dx} \]
\[ \Rightarrow 2I = 4\int\limits_0^{\dfrac{\pi }{2}} {\left[ {\log \sin 2x - \log 2} \right]dx} \]
\[ \Rightarrow 2I = 4\left[ {\int\limits_0^{\dfrac{\pi }{2}} {\log \sin 2xdx} - \int\limits_0^{\dfrac{\pi }{2}} {\log 2dx} } \right]\]
\[ \Rightarrow 2I = 4\left[ {\int\limits_0^{\dfrac{\pi }{2}} {\log \sin 2xdx} - \log 2\int\limits_0^{\dfrac{\pi }{2}} {dx} } \right]\] \[.....\left( 3 \right)\]
Now substitute \[2x = u\] in the first integral.
Then, \[dx = \dfrac{{du}}{2}\]
The limits changed as follows:
As \[x \to 0\], then \[u \to 0\]
As \[x \to \dfrac{\pi }{2}\], then \[u \to \pi \]

Substitute the values in the equation \[\left( 3 \right)\].
\[2I = 4\left[ {\dfrac{1}{2}\int\limits_0^\pi {\log \sin udu} - \log 2\int\limits_0^{\dfrac{\pi }{2}} {dx} } \right]\]
Apply the integration rule \[\int\limits_0^{2a} {f\left( x \right)} dx = 2\int\limits_0^a {f\left( x \right)} dx\] if \[f\left( x \right) = f\left( {2a - x} \right)\] on the first integral.
Here, \[\log \sin \left( {\pi - u} \right) = \log \sin u\] .
So, we get
\[ \Rightarrow 2I = 4\left[ {\dfrac{1}{2} \times 2\int\limits_0^{\dfrac{\pi }{2}} {\log \sin udu} - \log 2\left[ x \right]_0^{\dfrac{\pi }{2}}} \right]\]
\[ \Rightarrow 2I = 4\left[ {\int\limits_0^\pi {\log \sin udu} - \dfrac{\pi }{2}\log 2} \right]\]
\[ \Rightarrow 2I = 4\int\limits_0^\pi {\log \sin udu} - \dfrac{{4\pi }}{2}\log 2\]
From equation \[\left( 1 \right)\], we get \[I = 4\int\limits_0^\pi {\log \sin udu} \].
\[ \Rightarrow 2I = I - 2\pi \log 2\]
\[ \Rightarrow I = - 2\pi \log 2\]
\[ \Rightarrow I = 2\pi \log {2^{ - 1}}\]
\[ \Rightarrow I = 2\pi \log \left( {\dfrac{1}{2}} \right)\]
Thus, \[\int\limits_0^\pi {\log {{\sin }^2}xdx} = 2\pi \log \left( {\dfrac{1}{2}} \right)\].




Option ‘A’ is correct


Note: Students get confused and try to solve the integral \[\int {\log {{\sin }^2}xdx} \] by using the formula \[\int {\log x = x\left( {\log x - 1} \right)} \] . Because of that, they get the wrong answer.