# What is the value of integral int 10log[(1+x)1/2+(1−x)1/2]dx

What is the value of integral
${\int }_{0}^{1}\mathrm{log}\left[\left(1+x{\right)}^{1/2}+\left(1-x{\right)}^{1/2}\right]dx$
? I'm applying integration by parts but couldn't find any substitution after couple of steps
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amilazamiyn
Let
$I={\int }_{0}^{1}\mathrm{ln}\left(\sqrt{1+x}+\sqrt{1-x}\right)\cdot 1dx$
Integration by parts
$I=\mathrm{log}\left(\sqrt{1+x}+\sqrt{1-x}\right)\cdot x{|}_{0}^{1}+\frac{1}{2}{\int }_{0}^{1}\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\cdot \frac{1}{\left(\sqrt{1+x}\cdot \sqrt{1-x}\right)}xdx$
$I=\frac{\mathrm{ln}2}{2}+\frac{1}{2}{\int }_{0}^{1}\frac{{x}^{2}}{\left(1+\sqrt{1-{x}^{2}}\right)\cdot \sqrt{1-{x}^{2}}}dx$
Put $x=\mathrm{sin}\theta$ Then $dx=\mathrm{cos}\theta d\theta$
$I=\frac{\mathrm{ln}2}{2}+\frac{1}{2}{\int }_{0}^{\frac{\pi }{2}}\frac{{\mathrm{sin}}^{2}\theta \cdot \mathrm{cos}\theta }{\left(1+\mathrm{cos}\theta \right)\cdot \mathrm{cos}\theta }d\theta$
$I=\frac{\mathrm{ln}2}{2}+\frac{1}{2}{\int }_{0}^{\frac{\pi }{2}}\left(1-\mathrm{cos}\theta \right)d\theta =\frac{\mathrm{ln}2}{2}+\frac{\pi }{4}-\frac{1}{2}.$
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Aarav Atkins
If $x\in \left(0,1\right)$ we have $\sqrt{1-x}+\sqrt{1+x}=\sqrt{2+2\sqrt{1-{x}^{2}}}$, hence
${\int }_{0}^{1}\mathrm{log}\left(\sqrt{1+x}+\sqrt{1-x}\right)\phantom{\rule{thinmathspace}{0ex}}dx=\frac{1}{2}\left(\mathrm{log}2+{\int }_{0}^{1}\mathrm{log}\left(1+\sqrt{1-{x}^{2}}\right)\phantom{\rule{thinmathspace}{0ex}}dx\right).$
On the other hand,
$\begin{array}{rcl}{\int }_{0}^{1}\mathrm{log}\left(1+\sqrt{1-{x}^{2}}\right)\phantom{\rule{thinmathspace}{0ex}}dx& =& {\int }_{0}^{\pi /2}\mathrm{cos}\left(\theta \right)\mathrm{log}\left(1+\mathrm{cos}\theta \right)\phantom{\rule{thinmathspace}{0ex}}d\theta \\ \left(IBP\right)\phantom{\rule{1em}{0ex}}& =& {\int }_{0}^{\pi /2}\frac{{\mathrm{sin}}^{2}\theta }{1+\mathrm{cos}\theta }\phantom{\rule{thinmathspace}{0ex}}d\theta \\ & =& {\int }_{0}^{\pi /2}\left(1-\mathrm{cos}\theta \right)\phantom{\rule{thinmathspace}{0ex}}d\theta =\frac{\pi -2}{2}\end{array}$
hence:
${\int }_{0}^{1}\mathrm{log}\left(\sqrt{1+x}+\sqrt{1-x}\right)\phantom{\rule{thinmathspace}{0ex}}dx=\frac{\pi -2+2\mathrm{log}2}{4}.$