`int_(0)^(pi) log (1 +cos x) dx`=

Options

A. `pi "log" (1)/(2)`
B. `(pi)/(2) log 2`
C. `-pi log 2`
D. None

Answer

To solve the integral \( I = \int_0^{\pi} \log(1 + \cos x) \, dx \), we will use the property of definite integrals and some logarithmic identities. Here’s the step-by-step solution:

### Step 1: Define the Integral
Let
\[ I = \int_0^{\pi} \log(1 + \cos x) \, dx \]

### Step 2: Use the Property of Definite Integrals
We can use the property that states:
\[ \int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx \]
In our case, we set \( a = \pi \):
\[ I = \int_0^{\pi} \log(1 + \cos(\pi - x)) \, dx \]
Since \( \cos(\pi - x) = -\cos x \), we have:
\[ I = \int_0^{\pi} \log(1 - \cos x) \, dx \]

### Step 3: Add the Two Integrals
Now we add both representations of \( I \):
\[ 2I = \int_0^{\pi} \log(1 + \cos x) \, dx + \int_0^{\pi} \log(1 - \cos x) \, dx \]
Using the logarithmic property \( \log a + \log b = \log(ab) \):
\[ 2I = \int_0^{\pi} \log((1 + \cos x)(1 - \cos x)) \, dx \]

### Step 4: Simplify the Expression
Now, simplify the expression inside the logarithm:
\[ (1 + \cos x)(1 - \cos x) = 1 - \cos^2 x = \sin^2 x \]
Thus, we have:
\[ 2I = \int_0^{\pi} \log(\sin^2 x) \, dx \]

### Step 5: Use the Logarithmic Identity
We can use the identity \( \log(a^b) = b \log a \):
\[ 2I = 2 \int_0^{\pi} \log(\sin x) \, dx \]
Dividing both sides by 2:
\[ I = \int_0^{\pi} \log(\sin x) \, dx \]

### Step 6: Evaluate the Integral
We know from a standard result that:
\[ \int_0^{\pi} \log(\sin x) \, dx = -\frac{\pi}{2} \log 2 \]
Thus,
\[ I = -\frac{\pi}{2} \log 2 \]

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