Question

What is the value of \[\int_{ - \dfrac{1}{2}}^{\dfrac{1}{2}} {\cos x\ln \dfrac{{1 + x}}{{1 - x}}dx} \]?
A. 0
B. 1
C. 2
D. \[\ln 3\]

Answer 1

Hint: The given integration is a definite integral. Then we will check whether the function is an odd function or an even function. Then apply definite integration property to solve it.



Formula Used:Property of the definite integral:
\[\int\limits_{ - a}^a {f\left( x \right)dx} = \left\{ {\begin{array}{*{20}{c}}0&{{\rm{if f(x) is an odd function}}}\\{2\int_0^a {f\left( x \right)dx} }&{{\rm{if f(x) is an even function}}}\end{array}} \right.\]
Logarithm Rule:
\[\ln {a^b} = b\ln a\]



Complete step by step solution:Given definite integral is \[\int_{ - \dfrac{1}{2}}^{\dfrac{1}{2}} {\cos x\ln \dfrac{{1 + x}}{{1 - x}}dx} \].
Assume that, \[f\left( x \right) = \cos x\ln \dfrac{{1 + x}}{{1 - x}}\]
To check whether the given function is an odd function or even function, we will put x = -x in \[f\left( x \right)\].
\[f\left( { - x} \right) = \cos \left( { - x} \right)\ln \dfrac{{1 + \left( { - x} \right)}}{{1 - \left( { - x} \right)}}\]
\[ \Rightarrow f\left( { - x} \right) = \cos \left( { - x} \right)\ln \dfrac{{1 - x}}{{1 + x}}\]
We know that \[\cos \left( { - x} \right) = \cos x\].
\[ \Rightarrow f\left( { - x} \right) = \cos x\ln {\left( {\dfrac{{1 + x}}{{1 - x}}} \right)^{ - 1}}\]
Now applying the formula \[\ln {a^b} = b\ln a\]
\[ \Rightarrow f\left( { - x} \right) = - \cos x\ln \dfrac{{1 + x}}{{1 - x}}\]
\[ \Rightarrow f\left( { - x} \right) = - f\left( x \right)\]
Thus \[f\left( x \right)\] is an odd function.
Now we will apply the definite integral property \[\int\limits_{ - a}^a {f\left( x \right)dx} = \left\{ {\begin{array}{*{20}{c}}0&{{\rm{if f(x) is an odd function}}}\\{2\int_0^a {f\left( x \right)dx} }&{{\rm{if f(x) is an even function}}}\end{array}} \right.\].
Since \[\cos x\ln \dfrac{{1 + x}}{{1 - x}}\] is a odd function, then
\[\int_{ - \dfrac{1}{2}}^{\dfrac{1}{2}} {\cos x\ln \dfrac{{1 + x}}{{1 - x}}dx} = 0\]



Option ‘A’ is correct



Note: Students often make mistake when they put the value of \[\cos \left( { - x} \right)\]. They use \[\cos \left( { - x} \right) = - \cos x\] which is an incorrect formula. Since cosine function is an even function thus \[\cos \left( { - x} \right) = \cos x\].