Question

${\int }_{-\pi /2}^{\pi /2}\cos x$ ln $\left(\frac{1+x}{1-x}\right)dx$ is equal to.

• A. $0$
• B. $\frac{{\pi }^2}{4}(-1+\frac{\pi }{2})$
• C. $1$
• D. $\frac{{\pi }^2}{2}$

Answer 1

Answer: (A)

$0$

Let the given integral be $I$

We have $I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\cos x\ln \left(\frac{1+x}{1-x}\right)dx$

In the above expression , $x$ can be replaced with $\frac{\pi }{2}-\frac{\pi }{2}-x=-x$

$\Rightarrow I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\cos x\ln \left(\frac{1+x}{1-x}\right)dx={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\cos x\ln \left(\frac{1-x}{1+x}\right)dx$

By adding both, we get $2I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\cos x(\ln \left(\frac{1+x}{1-x}\right)+\ln \left(\frac{1-x}{1+x}\right))dx$

$={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\cos x\left(\ln (\frac{1+x}{1-x}\times \frac{1-x}{1+x})\right)dx={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\cos x\left(\ln \right(1\left)\right)dx=0$

$\Rightarrow I=0$

Therefore the correct option is $A$