Question

The value of ${\int }_{-\pi /2}^{\pi /2}\sin \left\{\log \right(x+\sqrt{x^2+1}\}dx$ is

• A. 1
• B. -1
• C. 0
• D. 2

Answer 1

Answer: (C)

0

Let $I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}sin\left(\log \sqrt{x^2+1}+x\right)dx$ (1)
Using property $I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}sin\left(\log \sqrt{x^2+1}+x\right)dx$
We get
$I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}sin\left(\log \sqrt{{(-x)}^2+1}-x\right)dx={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}sin\left(\log \left(\frac{1}{\sqrt{{\left(x\right)}^2+1}+x}\right)\right)dx$
$={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}sin\left(\log {(\sqrt{{\left(x\right)}^2+1}+x)}^{-1}\right)dx$
$=-{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}sin\left(\log (\sqrt{{\left(x\right)}^2+1}+x)\right)dx$ ...(2)
Adding (1) and (2), we get
$2I=0\Rightarrow I=0$