<s>[INST] Evaluation of $ \int_{0}^{1}\ln\left(\sqrt{1+x}-\sqrt{1-x}\right)dx$ 
Evaluation of $\displaystyle \int_{0}^{1}\ln\left(\sqrt{1+x}+\sqrt{1-x}\right)dx$

$\bf{My\; Try::}$ Let $$I = \frac{1}{2}\int_{0}^{1}\ln\left(\sqrt{1+x}+\sqrt{1-x}\right)^2dx = \frac{1}{2}\int_{0}^{1}\ln\left(2+2\sqrt{1-x^2}\right)dx$$
So $$I =\frac{1}{2}\ln(2)+\frac{1}{2}\int_{0}^{1}\ln\left(1+\sqrt{1-x^2}\right)\cdot 1dx$$
So $$I = \frac{\ln 2}{2}+\frac{1}{2}\left[x\cdot \ln(1+\sqrt{1-x^2})\right]_{0}^{1}+\frac{1}{2}\int_{0}^{1}\left(\frac{x^2}{1+\sqrt{1-x^2}}\right)\cdot\frac{1}{\sqrt{1-x^2}}dx$$
Now Put $x=\sin \phi\;,$ Then $dx = \cos \phi d\phi$ and changing limits, We get
So $$I = \frac{\ln 2}{2}+\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\left(1-\cos \phi\right) d\phi = \frac{\ln 2}{2}+\frac{\pi}{4}-\frac{1}{2}$$
Can we solve it any other way, If yes Then plz explain here
Thanks 
 [/INST] If $y=\ln\left(\sqrt{1+x}+\sqrt{1-x}\right)$, then after solving for $x$ while assuming $x$ is positive, you have  $x=e^y\sqrt{1-\frac{e^{2y}}{4}}$.
View the integral as an area. There is a rectangle that is $[0,1]\times[0,\ln\sqrt{2}]$, plus a section above that we can integrate with respect to $y$:

$$\int_{y=\ln\sqrt{2}}^{y=\ln2}e^y\sqrt{1-\frac{e^{2y}}{4}}\,dy$$ which is a candidate for $u$-substitutions with $u=e^{y}/2$:
$$\int_{u=\sqrt{2}/2}^{u=1}2\sqrt{1-u^2}\,du=\left[u\sqrt{1-u^2}+\arcsin(u)\right]_{u=\sqrt{2}/2}^{u=1}=\frac{\pi}{4}-\frac{1}{2}$$
Add this to the rectangle's area of $\ln\sqrt{2}$ to get $$\ln\sqrt{2}+\frac{\pi}{4}-\frac{1}{2}$$
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