Question

Evaluate : ${\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\cos x\log \left[\frac{\left(1+x\right)}{\left(1-x\right)}\right]dx$

• A. $0$
• B. $\frac{{\mathrm{\pi }}^2}{4}\left[1+\left(\frac{\mathrm{\pi }}{2}\right)\right]$
• C. $1$
• D. $\frac{{\mathrm{\pi }}^2}{2}$

Answer 1

The correct option is A

$0$

Evaluating the integral.

we know that ${\int }_{-a}^{a}f\left(x\right)dx=\left\{\begin{array}{l}0,\mathrm{f}\left(\mathrm{x}\right)\mathrm{is}\mathrm{odd}\mathrm{function}\\ 2{\int }_a^{a}f\left(x\right)dx,\mathrm{f}\left(\mathrm{x}\right)\mathrm{is}\mathrm{even}\mathrm{function}\end{array}\right.$

$f\left(x\right)$ is an odd function if $f\left(-x\right)=-f\left(x\right)$

$f\left(x\right)$is an even function if $f\left(-x\right)=f\left(x\right)$

As per the question $f\left(x\right)=\cos x\log \left(\frac{1+x}{1-x}\right)$

$\Rightarrow$ $f\left(-x\right)=\cos \left(-x\right)\log \left(\frac{1+\left(-x\right)}{1-\left(-x\right)}\right)$

$\Rightarrow$ $f\left(-x\right)=\cos \left(x\right)\log \left(\frac{1-x}{1+x}\right)$

$\Rightarrow$ $f\left(-x\right)=\cos \left(x\right){\log \left(\frac{1+x}{1-x}\right)}^{-1}$

$\Rightarrow$ $f\left(-x\right)=-\cos \left(x\right)\log \left(\frac{1+x}{1-x}\right)$

$\Rightarrow$ $f\left(-x\right)=-f\left(x\right)$

Hence $f\left(x\right)=\cos x\log \left(\frac{1+x}{1-x}\right)$ is an odd function.

Therefore, the value of ${\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}cosx\log \left[\frac{\left(1+x\right)}{\left(1-x\right)}\right]dx\mathrm{is}0$

Hence, option A is the correct answer