If `f(x)` is an even function, then what is `int_(0)^(pi) f(cos x)` dx equal to ?

Options

A. `0`
B. `int_(0)^(pi//2) f(cosx) dx`
C. `2int_(0)^(pi/2) f(cosx ) dx`
D. `1`

Explanation

To solve the integral \( \int_{0}^{\pi} f(\cos x) \, dx \) where \( f(x) \) is an even function, we can follow these steps:

### Step 1: Understanding the Even Function
Since \( f(x) \) is an even function, we know that:
\[ f(-x) = f(x) \]

### Step 2: Change of Variable
We can use the property of cosine to change the variable in the integral. Let’s perform a substitution:
\[ u = \pi - x \quad \Rightarrow \quad du = -dx \]
When \( x = 0 \), \( u = \pi \) and when \( x = \pi \), \( u = 0 \). Thus, the integral becomes:
\[ \int_{0}^{\pi} f(\cos x) \, dx = \int_{\pi}^{0} f(\cos(\pi - u)) (-du) = \int_{0}^{\pi} f(-\cos u) \, du \]

### Step 3: Using the Even Function Property
Since \( f \) is even, we have:
\[ f(-\cos u) = f(\cos u) \]
Thus, we can rewrite the integral as:
\[ \int_{0}^{\pi} f(\cos x) \, dx = \int_{0}^{\pi} f(\cos u) \, du \]

### Step 4: Combining the Integrals
Now we can add the two integrals:
\[ 2 \int_{0}^{\pi} f(\cos x) \, dx = \int_{0}^{\pi} f(\cos x) \, dx + \int_{0}^{\pi} f(-\cos x) \, dx \]
This implies:
\[ 2 \int_{0}^{\pi} f(\cos x) \, dx = 2 \int_{0}^{\frac{\pi}{2}} f(\cos x) \, dx \]
Thus, we can simplify:
\[ \int_{0}^{\pi} f(\cos x) \, dx = 2 \int_{0}^{\frac{\pi}{2}} f(\cos x) \, dx \]

### Final Result
Therefore, the final result is:
\[ \int_{0}^{\pi} f(\cos x) \, dx = 2 \int_{0}^{\frac{\pi}{2}} f(\cos x) \, dx \]