Question

Evaluate the value of \[\int_0^{\dfrac{\pi }{2}} {\log \sin xd} x\] .
A. \[ - \pi \log 2\]
B. \[\pi \log 2\]
C. \[ - \dfrac{\pi }{2}\log 2\]
D. \[\dfrac{\pi }{2}\log 2\]

Answer 1

Hint: The given question belongs to definite integral. By applying the below property of definite integral, we will solve the given integration.

Formula used:
Trigonometry ratios for complementary angle
\[\sin \left( {\dfrac{\pi }{2} - \theta } \right) = \cos \theta \]
Property of definite integral:
1. \[\int_0^a {f\left( x \right)dx} = \int_0^a {f\left( {a - x} \right)dx} \]
2. \[\int_0^{2a} {f\left( x \right)dx} = 2\begin{array}{*{20}{c}}{\int_0^a {f\left( x \right)dx} }&{i{\rm{f}}\,f\left( {2a - x} \right) = f\left( x \right)}\end{array}\]

Complete step by step solution:
Let \[I = \int_0^{\dfrac{\pi }{2}} {\log \sin xd} x\] …..(1)
Using the property 1 in the given integration
\[ \Rightarrow I = \int_0^{\dfrac{\pi }{2}} {\log \sin \left( {\dfrac{\pi }{2} - x} \right)d} x\]
We know that \[\sin \left( {\dfrac{\pi }{2} - \theta } \right) = \cos \theta \]. Applying complementary formula
\[ \Rightarrow I = \int_0^{\dfrac{\pi }{2}} {\log \cos xd} x\] ….(2)
Now we will add equation (1) and (2)
\[I + I = \int_0^{\dfrac{\pi }{2}} {\log \sin xd} x + \int_0^{\dfrac{\pi }{2}} {\log \cos xd} x\]
\[ \Rightarrow I + I = \int_0^{\dfrac{\pi }{2}} {\left( {\log \sin x + \log \cos x} \right)d} x\]
Now applying the product rule of logarithm
 \[ \Rightarrow I + I = \int_0^{\dfrac{\pi }{2}} {\log \left( {\sin x\cos x} \right)d} x\] [Since \[\log a + \log b = \log ab\]]
\[ \Rightarrow I + I = \int_0^{\dfrac{\pi }{2}} {\log \left( {\dfrac{{2\sin x\cos x}}{2}} \right)d} x\]
Now applying double angle formula
\[ \Rightarrow I + I = \int_0^{\dfrac{\pi }{2}} {\log \left( {\dfrac{{\sin 2x}}{2}} \right)d} x\] [ Since \[2\sin x\cos x = \sin 2x\]]
Applying the quotient rule of logarithm
\[ \Rightarrow 2I = \int_0^{\dfrac{\pi }{2}} {\log \sin 2xd} x - \int_0^{\dfrac{\pi }{2}} {\log 2dx} \] [Since \[\log a - \log b = \log \dfrac{a}{b}\]]
Let \[{I_1} = \int_0^{\dfrac{\pi }{2}} {\log \sin 2xd} x\].
Now we will solve \[{I_1}\].
Assume that \[2x = t\]
Differentiate both sides
\[2dx = dt\]
Changing the limits
If x = 0 then t = 0
If \[x = \dfrac{\pi}{2} \] then \[ t = \pi \]
Now substitute \[2dx = dt\] and \[2x = t\] in \[{I_1} = \int_0^{\dfrac{\pi }{2}} {\log \sin 2xd} x\]
\[{I_1} = \int_0^\pi {\log \sin t\dfrac{{dt}}{2}} \]
\[ \Rightarrow {I_1} = \dfrac{1}{2}\int_0^\pi {\log \sin tdt} \]
We know that \[\sin \theta = \sin \left( {\pi - \theta } \right)\].
Since \[\log \sin t = \log \sin \left( {\pi - t} \right)\], so we can apply the formula \[\int_0^{2a} {f\left( x \right)dx} = 2\begin{array}{*{20}{c}}{\int_0^a {f\left( x \right)dx} }&{i{\rm{f}}\,f\left( {2a - x} \right) = f\left( x \right)}\end{array}\].
\[{I_1} = \dfrac{1}{2} \times 2\int_0^{\dfrac{\pi }{2}} {\log \sin td} t\]
\[ \Rightarrow {I_1} = \int_0^{\dfrac{\pi }{2}} {\log \sin td} t\]
\[ \Rightarrow {I_1} = \int_0^{\dfrac{\pi }{2}} {\log \sin xd} x\] [Since \[\int_b^a {f\left( x \right)d} x = \int_b^a {f\left( t \right)d} t\]]
\[ \Rightarrow {I_1} = I\]
\[ \Rightarrow \int_0^{\dfrac{\pi }{2}} {\log \sin 2xd} x = I\] [Since \[{I_1} = \int_0^{\dfrac{\pi }{2}} {\log \sin 2xd} x\]]
Putting \[\int_0^{\dfrac{\pi }{2}} {\log \sin 2xd} x = I\] in the equation \[2I = \int_0^{\dfrac{\pi }{2}} {\log \sin 2xd} x - \int_0^{\dfrac{\pi }{2}} {\log 2dx} \]
\[2I = I - \int_0^{\dfrac{\pi }{2}} {\log 2dx} \]
\[ \Rightarrow 2I - I = - \log 2\int_0^{\dfrac{\pi }{2}} {dx} \]
\[ \Rightarrow 2I - I = - \log 2\left[ x \right]_0^{\dfrac{\pi }{2}}\]
\[ \Rightarrow I = - \log 2\left[ {\dfrac{\pi }{2} - 0} \right]\]
\[ \Rightarrow I = - \dfrac{\pi }{2}\log 2\]
Hence option C is the correct option

Note: Many students often confused with the formulas \[\int_0^{2a} {f\left( x \right)dx} = 2\begin{array}{*{20}{c}}{\int_0^a {f\left( x \right)dx} }&{i{\rm{f}}\,f\left( {2a - x} \right) = f\left( x \right)}\end{array}\] and \[\int_0^{2a} {f\left( x \right)dx} = \begin{array}{*{20}{c}}0&{i{\rm{f}}\,f\left( {2a - x} \right) = f\left( x \right)}\end{array}\]. The correct formulas are \[\int_0^{2a} {f\left( x \right)dx} = 2\begin{array}{*{20}{c}}{\int_0^a {f\left( x \right)dx} }&{i{\rm{f}}\,f\left( {2a - x} \right) = f\left( x \right)}\end{array}\] and \[\int_0^{2a} {f\left( x \right)dx} = 2\begin{array}{*{20}{c}}{\int_0^a {f\left( x \right)dx} }&{i{\rm{f}}\,f\left( {2a - x} \right) = - f\left( x \right)}\end{array}\].