Evaluate the following definite integrals:
`int_(0)^(pi//2)(1-sinx)/(x+cos x)dx`

Answer

To evaluate the definite integral

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{1 - \sin x}{x + \cos x} \, dx, \]

we will use a substitution method.

### Step 1: Substitution
Let's make the substitution \( t = x + \cos x \).

To find \( dt \), we differentiate \( t \):
\[ dt = (1 - \sin x) \, dx. \]

### Step 2: Change of Limits
Now we need to change the limits of integration.

- When \( x = 0 \):
\[ t = 0 + \cos(0) = 1. \]

- When \( x = \frac{\pi}{2} \):
\[ t = \frac{\pi}{2} + \cos\left(\frac{\pi}{2}\right) = \frac{\pi}{2} + 0 = \frac{\pi}{2}. \]

So, the limits of integration change from \( x = 0 \) to \( x = \frac{\pi}{2} \) into \( t = 1 \) to \( t = \frac{\pi}{2} \).

### Step 3: Rewrite the Integral
Now we can rewrite the integral in terms of \( t \):
\[ I = \int_{1}^{\frac{\pi}{2}} \frac{1}{t} \, dt. \]

### Step 4: Evaluate the Integral
The integral of \( \frac{1}{t} \) is:
\[ \int \frac{1}{t} \, dt = \log t. \]

Thus, we have:
\[ I = \left[ \log t \right]_{1}^{\frac{\pi}{2}} = \log\left(\frac{\pi}{2}\right) - \log(1). \]

Since \( \log(1) = 0 \), we get:
\[ I = \log\left(\frac{\pi}{2}\right). \]

### Final Answer
The value of the integral is:
\[ \boxed{\log\left(\frac{\pi}{2}\right)}. \]