Solve
${\sin }^{-1}\left[\frac{\sqrt{1+x}+\sqrt{1-x}}{2}\right]$

We have,

${\sin }^{-1}\left(\frac{\sqrt{1+x}+\sqrt{1-x}}{2}\right)$

Let put $x=\cos 2\theta$

So,

${\sin }^{-1}\left(\frac{\sqrt{1+\cos 2\theta }+\sqrt{1-\cos 2\theta }}{2}\right)$

$\Rightarrow {\sin }^{-1}\left(\frac{\sqrt{1+2{\cos }^2\theta -1}+\sqrt{1-1+2{\sin }^2\theta }}{2}\right)$

$\Rightarrow {\sin }^{-1}\left(\frac{\sqrt{2{\cos }^2\theta }+\sqrt{2{\sin }^2\theta }}{2}\right)$

$\Rightarrow {\sin }^{-1}\left(\frac{\cos \theta +\sin \theta }{\sqrt{2}}\right)$

$\Rightarrow {\sin }^{-1}\left(\frac{\cos \theta }{\sqrt{2}}+\frac{\sin \theta }{\sqrt{2}}\right)$

$\Rightarrow {\sin }^{-1}\left(sin{45}^o\cos \theta +cos{45}^o\sin \theta \right)$

$\Rightarrow {\sin }^{-1}\sin \left({45}^o+\theta \right)$

$\Rightarrow {45}^o+\theta$

$\Rightarrow {45}^o+\frac{1}{2}{\cos }^{-1}x$

Hence, this is the answer.