Problem
Simplify: \( \sin^{-1}\left(\frac{\sqrt{1+x} + \sqrt{1-x}}{2}\right), \quad 0 < x < 1 \)
Solution (Substitution Method)
Let:
\[ x = \cos 2\theta \]
Then,
\[ 1 + x = 1 + \cos 2\theta = 2\cos^2 \theta \]
\[ 1 – x = 1 – \cos 2\theta = 2\sin^2 \theta \]
So,
\[ \sqrt{1+x} = \sqrt{2}\cos \theta, \quad \sqrt{1-x} = \sqrt{2}\sin \theta \]
Thus the expression becomes:
\[ \sin^{-1}\left(\frac{\sqrt{2}\cos \theta + \sqrt{2}\sin \theta}{2}\right) \]
\[ = \sin^{-1}\left(\frac{\cos \theta + \sin \theta}{\sqrt{2}}\right) \]
Using identity:
\[ \cos \theta + \sin \theta = \sqrt{2}\sin\left(\theta + \frac{\pi}{4}\right) \]
So,
\[ \frac{\cos \theta + \sin \theta}{\sqrt{2}} = \sin\left(\theta + \frac{\pi}{4}\right) \]
Hence,
\[ \sin^{-1}\left(\sin\left(\theta + \frac{\pi}{4}\right)\right) = \theta + \frac{\pi}{4} \]
Since \( x = \cos 2\theta \), we get:
\[ 2\theta = \cos^{-1} x \quad \Rightarrow \quad \theta = \frac{1}{2}\cos^{-1} x \]
Final Answer
\[ \boxed{\frac{1}{2}\cos^{-1}x + \frac{\pi}{4}} \]