Evaluate: `int_0^pi log(1+cosx)dx`

Explanation

To evaluate the integral \( I = \int_0^{\pi} \log(1 + \cos x) \, dx \), we can use a property of definite integrals. We will first rewrite the integral using the substitution \( x = \pi - t \).

### Step-by-step Solution:

1. **Rewrite the Integral**:
\[ I = \int_0^{\pi} \log(1 + \cos x) \, dx \]
Using the substitution \( x = \pi - t \), we have \( dx = -dt \). Changing the limits accordingly, when \( x = 0 \), \( t = \pi \) and when \( x = \pi \), \( t = 0 \):
\[ I = \int_{\pi}^{0} \log(1 + \cos(\pi - t)) (-dt) = \int_0^{\pi} \log(1 - \cos t) \, dt \]

2. **Combine the Integrals**:
Now we have:
\[ I = \int_0^{\pi} \log(1 + \cos x) \, dx = \int_0^{\pi} \log(1 - \cos x) \, dx \]
Adding these two equations:
\[ 2I = \int_0^{\pi} \log(1 + \cos x) \, dx + \int_0^{\pi} \log(1 - \cos x) \, dx \]

3. **Use Logarithmic Properties**:
We can combine the logarithms:
\[ 2I = \int_0^{\pi} \log((1 + \cos x)(1 - \cos x)) \, dx = \int_0^{\pi} \log(\sin^2 x) \, dx \]
Since \( 1 - \cos^2 x = \sin^2 x \).

4. **Simplify the Integral**:
The integral simplifies to:
\[ 2I = \int_0^{\pi} 2 \log(\sin x) \, dx \]
Thus:
\[ I = \int_0^{\pi} \log(\sin x) \, dx \]

5. **Evaluate the Integral**:
The integral \( \int_0^{\pi} \log(\sin x) \, dx \) is a known result:
\[ \int_0^{\pi} \log(\sin x) \, dx = -\pi \log(2) \]
Therefore:
\[ I = -\frac{\pi}{2} \log(2) \]

6. **Final Result**:
Thus, the value of the integral \( \int_0^{\pi} \log(1 + \cos x) \, dx \) is:
\[ I = -\frac{\pi}{2} \log(2) \]