The value of the integral `overset(pi)underset(0)int log(1+cos x)dx` is

Options

A. `(pi)/(2)log2`
B. `-pi log2`
C. `pi log2`
D. none of these

Answer

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

### Step 1: Set up the integral
We start with the integral:
\[ I = \int_0^{\pi} \log(1 + \cos x) \, dx \]

### 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, let \( a = \pi \):
\[ I = \int_0^{\pi} \log(1 + \cos(\pi - x)) \, dx \]
Since \( \cos(\pi - x) = -\cos x \), we can rewrite the integral:
\[ I = \int_0^{\pi} \log(1 - \cos x) \, dx \]

### Step 3: Combine the two integrals
Now we have two expressions for \( I \):
\[ I = \int_0^{\pi} \log(1 + \cos x) \, dx \quad \text{(1)} \]
\[ I = \int_0^{\pi} \log(1 - \cos x) \, dx \quad \text{(2)} \]
Adding these two equations gives:
\[ 2I = \int_0^{\pi} \left( \log(1 + \cos x) + \log(1 - \cos x) \right) \, dx \]
Using the property of logarithms, we can combine the logs:
\[ 2I = \int_0^{\pi} \log((1 + \cos x)(1 - \cos x)) \, dx \]
This simplifies to:
\[ 2I = \int_0^{\pi} \log(1 - \cos^2 x) \, dx \]

### Step 4: Simplify the expression
Using the identity \( 1 - \cos^2 x = \sin^2 x \), we have:
\[ 2I = \int_0^{\pi} \log(\sin^2 x) \, dx \]
This can be further simplified:
\[ 2I = 2 \int_0^{\pi} \log(\sin x) \, dx \]
Thus,
\[ I = \int_0^{\pi} \log(\sin x) \, dx \]

### Step 5: Evaluate the integral
The integral \( \int_0^{\pi} \log(\sin x) \, dx \) is a known result and evaluates to:
\[ \int_0^{\pi} \log(\sin x) \, dx = -\pi \log(2) \]
Therefore, we have:
\[ I = -\pi \log(2) \]

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