Evaluate: `int_0^pi |cos x| dx`

Answer

To evaluate the integral \( I = \int_0^\pi |\cos x| \, dx \), we need to analyze the behavior of the function \( |\cos x| \) over the interval from \( 0 \) to \( \pi \).

### Step 1: Identify the points where \( \cos x \) changes sign
The cosine function is positive in the interval \( [0, \frac{\pi}{2}] \) and negative in the interval \( [\frac{\pi}{2}, \pi] \). Therefore, we can break the integral into two parts:
\[ I = \int_0^{\frac{\pi}{2}} |\cos x| \, dx + \int_{\frac{\pi}{2}}^{\pi} |\cos x| \, dx \]

### Step 2: Simplify the absolute value
In the first interval \( [0, \frac{\pi}{2}] \), \( \cos x \) is non-negative, so \( |\cos x| = \cos x \). In the second interval \( [\frac{\pi}{2}, \pi] \), \( \cos x \) is negative, so \( |\cos x| = -\cos x \). Thus, we can rewrite the integral as:
\[ I = \int_0^{\frac{\pi}{2}} \cos x \, dx + \int_{\frac{\pi}{2}}^{\pi} -\cos x \, dx \]

### Step 3: Evaluate the first integral
Now we evaluate the first integral:
\[ \int_0^{\frac{\pi}{2}} \cos x \, dx \]
The antiderivative of \( \cos x \) is \( \sin x \). Therefore,
\[ \int_0^{\frac{\pi}{2}} \cos x \, dx = \left[ \sin x \right]_0^{\frac{\pi}{2}} = \sin\left(\frac{\pi}{2}\right) - \sin(0) = 1 - 0 = 1 \]

### Step 4: Evaluate the second integral
Next, we evaluate the second integral:
\[ \int_{\frac{\pi}{2}}^{\pi} -\cos x \, dx \]
This can be rewritten as:
\[ -\int_{\frac{\pi}{2}}^{\pi} \cos x \, dx \]
Again, using the antiderivative of \( \cos x \):
\[ -\int_{\frac{\pi}{2}}^{\pi} \cos x \, dx = -\left[ \sin x \right]_{\frac{\pi}{2}}^{\pi} = -\left( \sin(\pi) - \sin\left(\frac{\pi}{2}\right) \right) = -\left( 0 - 1 \right) = 1 \]

### Step 5: Combine the results
Now we can combine the results of both integrals:
\[ I = 1 + 1 = 2 \]

Thus, the value of the integral is:
\[ \boxed{2} \]