We need to evaluate \(\int\sqrt{1+\sin x}dx\).

We can rewrite \(1+\sin x\) using trigonometric identities:

\(1+\sin x = \sin^2\frac{x}{2} + \cos^2\frac{x}{2} + 2\sin\frac{x}{2}\cos\frac{x}{2} = (\sin\frac{x}{2} + \cos\frac{x}{2})^2\)

So, \(\sqrt{1+\sin x} = \sqrt{(\sin\frac{x}{2} + \cos\frac{x}{2})^2} = |\sin\frac{x}{2} + \cos\frac{x}{2}|\)

Assuming \(\sin\frac{x}{2} + \cos\frac{x}{2} > 0\), we have:

\(\int\sqrt{1+\sin x}dx = \int (\sin\frac{x}{2} + \cos\frac{x}{2})dx\)

\(= \int \sin\frac{x}{2}dx + \int \cos\frac{x}{2}dx\)

\(= -2\cos\frac{x}{2} + 2\sin\frac{x}{2} + C\)

\(= 2(\sin\frac{x}{2} - \cos\frac{x}{2}) + C\)