Question

Evaluate ${\int }_0^1\log [\sqrt{1-x}+\sqrt{1+x}]dx=\frac{1}{2}\log 2+\frac{\pi }{k}-\frac{1}{2}$

Answer 1

Let $I={\int }_0^1\log [\sqrt{1-x}+\sqrt{1+x}]dx$
$={\left[x\log \{\sqrt{(1-x)}+\sqrt{(1+x)}\}\right]}_0^1-{\int }_0^{1}x.\frac{1}{\sqrt{(1-x)}+\sqrt{(1+x)}}.\{\frac{-1}{2\sqrt{(1-x)}}+\frac{1}{2\sqrt{(1+x)}}\}dx$
$=\log \sqrt{2}-{\int }_0^1\frac{x}{2\sqrt{(1-x^2)}}\frac{\sqrt{(1-x)}-\sqrt{(1+x)}}{\sqrt{(1-x)}+\sqrt{(1+x)}}dx$
Rationalize
$=\log \sqrt{2}-{\int }_0^1\frac{x}{2\sqrt{(1-x^2)}}\frac{{[\sqrt{(1-x)}-\sqrt{(1+x)}]}^2}{(1-x)-(1+x)}dx$
$=\log \sqrt{2}-{\int }_0^1\frac{x}{2\sqrt{(1-x^2)}}.\frac{(1-x)+(1+x)-2\sqrt{(1-x^2)}}{-2x}dx$
$=\log \sqrt{2}+{\int }_0^1\frac{1}{4\sqrt{(1-x^2)}}\{2-2\sqrt{(1-x^2)}\}dx$
$=\log \sqrt{2}+\frac{1}{2}{[{\sin }^{-1}x-x]}_0^1=\log \sqrt{2}+\frac{1}{2}[\frac{\pi }{2}-1]=\frac{1}{2}\log 2+\frac{\pi }{4}-\frac{1}{2}$