Question

What is the value of the definite integral \[\int\limits_0^{\dfrac{\pi }{2}} {x\cot x} dx\]?
A. \[ - \dfrac{\pi }{2}\log 2\]
B. \[\dfrac{\pi }{2}\log 2\]
C. \[\pi \log 2\]
D. \[ - \pi \log 2\]

Answer 1

Hint: Here, a definite integral is given. First, solve the integral by applying the integration by parts method. Then, solve the first term by applying the limits. After that, solve the simplify the integral by applying the integration rule \[\int\limits_0^a {f\left( x \right)} dx = \int\limits_0^a {f\left( {a - x} \right)} dx\]. 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:Integration by parts formula: \[\int {f\left( x \right)} g\left( x \right)dx = f\left( x \right)\int {g\left( x \right)} dx - \int {\left[ {\dfrac{d}{{dx}}f\left( x \right)\int {g\left( x \right)dx} } \right]} dx\]
\[\int\limits_0^{na} {f\left( x \right)} dx = n\int\limits_0^a {f\left( x \right)} dx\]
\[\int\limits_0^a {f\left( x \right)} dx = \int\limits_0^a {f\left( {a - x} \right)} dx\]
\[\log \left( a \right) + \log \left( b \right) = \log \left( {ab} \right)\]
\[\int\limits_a^b {{x^n}dx = \left[ {\dfrac{{{x^{n + 1}}}}{{n + 1}}} \right]} _a^b\]
\[\sin 2x = 2\sin x\cos x\]
\[\int {\cot xdx = \log \sin x} \]



Complete step by step solution:The given definite integral is \[\int\limits_0^{\dfrac{\pi }{2}} {x\cot x} dx\].
Let consider,
\[I = \int\limits_0^{\dfrac{\pi }{2}} {x\cot x} dx\]
Now solve the integral by the integration by parts formula \[\int {f\left( x \right)} g\left( x \right)dx = f\left( x \right)\int {g\left( x \right)} dx - \int {\left[ {\dfrac{d}{{dx}}f\left( x \right)\int {g\left( x \right)dx} } \right]} dx\]
\[I = \left[ {x\int {\cot xdx} } \right]_0^{\dfrac{\pi }{2}} - \int\limits_0^{\dfrac{\pi }{2}} {\left[ {\dfrac{d}{{dx}}x\int {\cot xdx} } \right]} dx\]
\[ \Rightarrow I = \left[ {x\log \sin x} \right]_0^{\dfrac{\pi }{2}} - \int\limits_0^{\dfrac{\pi }{2}} {\left[ {1 \times \log \sin x} \right]} dx\]
\[ \Rightarrow I = \left[ {\dfrac{\pi }{2}\log \sin \dfrac{\pi }{2} - 0\log \sin 0} \right] - \int\limits_0^{\dfrac{\pi }{2}} {\log \sin x} dx\]
\[ \Rightarrow I = \left[ {\dfrac{\pi }{2}\log 1 - 0} \right] - \int\limits_0^{\dfrac{\pi }{2}} {\log \sin x} dx\]
\[ \Rightarrow I = \dfrac{\pi }{2}\left( 0 \right) - \int\limits_0^{\dfrac{\pi }{2}} {\log \sin x} dx\]
\[ \Rightarrow I = - \int\limits_0^{\dfrac{\pi }{2}} {\log \sin x} dx\] \[.....\left( 1 \right)\]
Now apply the integration rule \[\int\limits_0^a {f\left( x \right)} dx = \int\limits_0^a {f\left( {a - x} \right)} dx\].
\[ \Rightarrow I = - \int\limits_0^{\dfrac{\pi }{2}} {\log \sin \left( {\dfrac{\pi }{2} - x} \right)dx} \]
\[ \Rightarrow I = - \int\limits_0^{\dfrac{\pi }{2}} {\log \cos xdx} \] \[.....\left( 2 \right)\]

Add the equations \[\left( 1 \right)\] and \[\left( 2 \right)\].
\[ \Rightarrow I + I = - \left[ {\int\limits_0^{\dfrac{\pi }{2}} {\log \sin xdx} + \int\limits_0^{\dfrac{\pi }{2}} {\log \cos xdx} } \right]\]
Apply the sum rule of integration \[\int\limits_a^b {\left[ {f\left( x \right) + g\left( x \right)} \right]dx} = \int\limits_a^b {f\left( x \right)dx} + \int\limits_a^b {g\left( x \right)dx} \].
\[ \Rightarrow 2I = - \int\limits_0^{\dfrac{\pi }{2}} {\left[ {\log \sin x + \log \cos x} \right]dx} \]
Apply the sum property of the logarithm \[\log \left( a \right) + \log \left( b \right) = \log \left( {ab} \right)\] .
\[ \Rightarrow 2I = - \int\limits_0^{\dfrac{\pi }{2}} {\left[ {\log \sin x\cos x} \right]dx} \]
\[ \Rightarrow 2I = - \int\limits_0^{\dfrac{\pi }{2}} {\log \left( {\dfrac{{2\sin x\cos x}}{2}} \right)dx} \]
\[ \Rightarrow 2I = - \int\limits_0^{\dfrac{\pi }{2}} {\log \left( {\dfrac{{\sin 2x}}{2}} \right)dx} \]
Apply the quotient property of logarithm \[\log \left( {\dfrac{a}{b}} \right) = \log \left( a \right) - \log \left( b \right)\]
\[ \Rightarrow 2I = - \int\limits_0^{\dfrac{\pi }{2}} {\log \left( {\dfrac{{\sin 2x}}{2}} \right)dx} \]
\[ \Rightarrow 2I = - \int\limits_0^{\dfrac{\pi }{2}} {\left[ {\log \sin 2x - \log 2} \right]dx} \]
\[ \Rightarrow 2I = \int\limits_0^{\dfrac{\pi }{2}} {\log 2dx} - \int\limits_0^{\dfrac{\pi }{2}} {\log \sin 2xdx} \]
\[ \Rightarrow 2I = \log 2\int\limits_0^{\dfrac{\pi }{2}} {dx} - \int\limits_0^{\dfrac{\pi }{2}} {\log \sin 2xdx} \] \[.....\left( 3 \right)\]
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 \[\left( 3 \right)\].
\[2I = \log 2\int\limits_0^{\dfrac{\pi }{2}} {dx} - \dfrac{1}{2}\int\limits_0^\pi {\log \sin udu} \]
Apply the integration rule \[\int\limits_0^{na} {f\left( x \right)} dx = n\int\limits_0^a {f\left( x \right)} dx\].
\[2I = \log 2\int\limits_0^{\dfrac{\pi }{2}} {dx} - \int\limits_0^{\dfrac{\pi }{2}} {\log \sin udu} \]
From equation \[\left( 1 \right)\], we get \[I = - \int\limits_0^{\dfrac{\pi }{2}} {\log \sin udu} \].
\[2I = \log 2\int\limits_0^{\dfrac{\pi }{2}} {dx} + I\]
\[ \Rightarrow I = \log 2\left[ x \right]_0^{\dfrac{\pi }{2}}\]
\[ \Rightarrow I = \log 2\left[ {\dfrac{\pi }{2} - 0} \right]\]
\[ \Rightarrow I = \dfrac{\pi }{2}\log 2\]
Thus, \[\int\limits_0^{\dfrac{\pi }{2}} {x\cos xdx} = \dfrac{\pi }{2}\log 2\]



Option ‘B’ 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.