The problem is to evaluate the definite integral $\int_{0}^{\frac{\pi}{2}} \frac{\cos x}{1 + \sin x} dx$.

1. Problem Description

The problem is to evaluate the definite integral $\int_{0}^{\frac{\pi}{2}} \frac{\cos x}{1 + \sin x} dx$.

2. Solution Steps

We can solve this integral using substitution. Let $u = 1 + \sin x$. Then, $\frac{du}{dx} = \cos x$, so $du = \cos x \, dx$.

When $x = 0$, $u = 1 + \sin 0 = 1 + 0 = 1$.

When $x = \frac{\pi}{2}$, $u = 1 + \sin \frac{\pi}{2} = 1 + 1 = 2$.

Thus, the integral becomes:

$\int_{1}^{2} \frac{1}{u} du$

The integral of $\frac{1}{u}$ is $\ln|u|$. So we have:

$\int_{1}^{2} \frac{1}{u} du = \ln|u| \Big|_1^2 = \ln|2| - \ln|1| = \ln 2 - \ln 1$

Since $\ln 1 = 0$, we have:

$\ln 2 - 0 = \ln 2$

3. Final Answer

$\ln{2}$