Answer
$$2\sqrt{2}$$
Work Step by Step
Given
$$ \int_{-\pi / 2}^{\pi / 2} \frac{\cos x d x}{(\sqrt{\sin x+1)}} $$
Let
$$ u= \sin x +1 \ \ \ \Rightarrow \ \ \ du =\cos x dx$$
At $$ x= -\pi/2\to u= 0, \ \ x= \pi/2\to u= 2$$
Then
\begin{aligned} \int_{-\pi / 2}^{\pi / 2} \frac{\cos x d x}{(\sqrt{\sin x+1)}} &=\int_{0}^{2} \frac{1}{\sqrt{u}} d u \\ &=\int_{0}^{2} u^{-1 / 2} d u \\ &=\left.2 u^{1 / 2}\right|_{0} ^{2} \\ &=2\sqrt{2} \end{aligned}
