Question

$\int\limits_{-1/2}^{1/2} \cos x\log\left(\frac{1+x}{1-x}\right) dx =$

• A. $0$
• B. $\sqrt{e}$
• C. $1$
• D. $2 \sqrt{e}$

Answer 1

Answer: (A)

$0$

Answer 2

$\int\limits_{-1 2}^{1 2} \cos x\log\left(\frac{1+x}{1-x}\right) dx =$
Let $f(x) = \cos x \log\left(\frac{1+x}{1-x}\right) = g(x) h(x)$
Here $g(x) = \cos x$ and $h(x) = \log \left(\frac{1+x}{1-x}\right) g(x)$ is an even function.
Now, $h(-x) = \log \left(\frac{1+x}{1-x}\right) = \log \left(\frac{1+x}{1-x}\right)^{-1}$
$= - \log \left(\frac{1+x}{1-x}\right) = - h(x) \: i.e. , h(x)$ is an odd function
$\therefore \, \, \, f(x)$ is an odd function