Question

The value of ${\int }_{-1/2}^{1/2}cosxln(x+\sqrt{1+x^2})dx$ is equal to ?

• A. $1$
• B. $0$
• C. $-In2$
• D. $\sqrt{2}+In2$

Answer 1

The correct option is B $0$

Consider $f\left(x\right)=\ln (x+\sqrt{1+x^2})$
Then,
$f(-x)=\ln (-x+\sqrt{1+x^2})$
$=-\ln (\frac{1}{\sqrt{1+x^2}-x})$
$=-\ln (\frac{1}{\sqrt{1+x^2}-x}\times \frac{\sqrt{1+x^2}+x}{\sqrt{1+x^2}+x})$
$=-\ln (\sqrt{1+x^2}+x)$
$=-f\left(x\right)$
Therefore, $f$ is an odd function.
This means that the product of $\cos x$ and $f\left(x\right)$ is an odd function as $\cos x$ is an even function.
Definite integral of an odd function from $-a$ to $a$ is always $0$.