Answer
$$
\int \sqrt{1-\sin x} d x =2 \sqrt{1+\sin x}+C
$$
where $C$ is an arbitrary constant.
Work Step by Step
$$
\int \sqrt{1-\sin x} d x
$$
To evaluate this integral we use the substitution as follows :
$$\begin{aligned}
\int \sqrt{1-\sin x} d x &=\int \sqrt{\frac{1-\sin x}{1} \cdot \frac{1+\sin x}{1+\sin x}} d x \\
&=\int \sqrt{\frac{1-\sin ^{2} x}{1+\sin x}} d x \\
&=\int \sqrt{\frac{\cos ^{2} x}{1+\sin x}} d x \\
&=\int \frac{\cos x d x}{\sqrt{1+\sin x}} \quad[\text { assume } \cos x>0] \\
& \quad\quad\quad\quad\left[\begin{array}{c}
u=1+\sin x, \\
d u=\cos x d x
\end{array}\right] \\
&=\int \frac{d u}{\sqrt{u}}\\
&=2 \sqrt{u}+C \\
&=2 \sqrt{1+\sin x}+C
\end{aligned}
$$
where $C$ is an arbitrary constant.