Answer
$$2-\sqrt{2}$$
Work Step by Step
We integrate as follows:
\begin{align*}
\int_{0}^{\pi / 6} \sqrt{1+\sin x} d x&=\int_{0}^{\pi / 6} \frac{\sqrt{1+\sin x}}{1} d x\\
&=\int_{0}^{\pi / 6} \frac{\sqrt{1-\sin ^{2} x}}{\sqrt{1-\sin x}} d x\\
&=\int_{0}^{\pi / 6} \frac{\sqrt{\cos ^{2} x}}{\sqrt{1-\sin x}} d x\\
&=\int_{0}^{\pi / 6} \frac{\cos x}{\sqrt{1-\sin x}} d x
\\
&= -2(1-\sin x)^{1 / 2}\bigg|_{0}^{\pi / 6}\\
&=-2 \sqrt{1-\sin \left(\frac{\pi}{6}\right)}+2 \sqrt{1-\sin 0}\\
&=-2 \sqrt{\frac{1}{2}}+2 \sqrt{1}\\
&=2-\sqrt{2}
\end{align*}
