Question

The value of the integral $\underset{-1}{\overset{1}{\int }}{\log }_e(\sqrt{1-x}+\sqrt{1+x})dx$ is equal to

• A. $2{\log }_{e}2+\frac{\pi }{4}-1$
• B. ${\log }_{e}2+\frac{\pi }{2}-1$
• C. $\frac{1}{2}{\log }_{e}2+\frac{\pi }{4}-\frac{3}{2}$
• D. $2{\log }_{e}2+\frac{\pi }{2}-\frac{1}{2}$

Answer 1

The correct option is B ${\log }_{e}2+\frac{\pi }{2}-1$

$I={\int }_{-1}^1{\log }_e(\sqrt{1-x}+\sqrt{1+x})dx$
Using property
If $f\left(x\right)$ is an even function, then ${\int }_{-a}^{a}f\left(x\right)dx=2{\int }_0^{a}f\left(x\right)dx$
$=2{\int }_0^1{\log }_e(\sqrt{1-x}+\sqrt{1+x})\cdot \left(1\right)dx$
$=2{[{\log }_e(\sqrt{1-x}+\sqrt{1+x})\cdot (x\left)\right]}_0^1-2{\int }_0^1\frac{1}{(\sqrt{1-x}+\sqrt{1+x})}\cdot (\frac{1}{2\sqrt{1+x}}-\frac{1}{2\sqrt{1-x}})\cdot xdx$
$=2({\log }_e\sqrt{2}-0)-\frac{2}{2}{\int }_0^{1}x\cdot \frac{\sqrt{1-x}-\sqrt{1+x}}{\sqrt{1-x}+\sqrt{1+x}}\times \frac{1}{\sqrt{1-x^2}}dx$
$={\log }_{e}2-{\int }_0^{1}x\cdot \frac{(\sqrt{1-x}-\sqrt{1+x}{)}^2}{1-x-1-x}\times \frac{1}{\sqrt{1-x^2}}dx$
$={\log }_{e}2-{\int }_0^{1}x\cdot \frac{1-x+1+x-2\sqrt{1-x^2}}{-2x}\times \frac{1}{\sqrt{1-x^2}}dx$
$={\log }_{e}2-{\int }_0^1\frac{2x(1-\sqrt{1-x^2})}{-2x}\times \frac{1}{\sqrt{1-x^2}}dx$
$={\log }_{e}2+{\int }_0^1\left(\frac{1-\sqrt{1-x^2}}{\sqrt{1-x^2}}\right)dx$
$={\log }_{e}2+{[{\sin }^{-1}x-x]}_0^1$
$={\log }_{e}2+[\frac{\pi }{2}-1-0]$
$I={\log }_{e}2+\frac{\pi }{2}-1$