Evaluate the following integrals:
`int_(0)^(pi) |cos x| dx`

Answer

To evaluate the integral \( \int_{0}^{\pi} |\cos x| \, dx \), we will follow these steps:

### Step 1: Identify the intervals where \( \cos x \) is positive and negative
The function \( \cos x \) is positive in the interval \( [0, \frac{\pi}{2}] \) and negative in the interval \( [\frac{\pi}{2}, \pi] \). Therefore, we can express the integral as a sum of two integrals:

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

### Step 2: Evaluate the first integral
For the first integral, since \( \cos x \) is positive in the interval \( [0, \frac{\pi}{2}] \):

\[ \int_{0}^{\frac{\pi}{2}} \cos x \, dx \]

The integral of \( \cos x \) is \( \sin x \). Thus, we evaluate:

\[ \int_{0}^{\frac{\pi}{2}} \cos x \, dx = \left[ \sin x \right]_{0}^{\frac{\pi}{2}} = \sin\left(\frac{\pi}{2}\right) - \sin(0) = 1 - 0 = 1 \]

### Step 3: Evaluate the second integral
For the second integral, since \( \cos x \) is negative in the interval \( [\frac{\pi}{2}, \pi] \):

\[ \int_{\frac{\pi}{2}}^{\pi} -\cos x \, dx \]

Again, the integral of \( -\cos x \) is \( -\sin x \). Thus, we evaluate:

\[ \int_{\frac{\pi}{2}}^{\pi} -\cos x \, dx = \left[-\sin x \right]_{\frac{\pi}{2}}^{\pi} = -\sin(\pi) - (-\sin\left(\frac{\pi}{2}\right)) = 0 - (-1) = 1 \]

### Step 4: Combine the results
Now, we combine the results of both integrals:

\[ \int_{0}^{\pi} |\cos x| \, dx = 1 + 1 = 2 \]

### Final Answer
Thus, the value of the integral \( \int_{0}^{\pi} |\cos x| \, dx \) is:

\[ \boxed{2} \]