If f(x) is an even function , then what is `int_0^pi f (cos x) dx` equal to

Options

A. A. 0
B. B. `int_0^(pi//2) f(cos x) dx`
C. C. `2 int_0^(pi//2) f (cos x) dx`
D. D. 1

Answer

To solve the problem, we need to evaluate the integral \( \int_0^\pi f(\cos x) \, dx \) given that \( f(x) \) is an even function.

### Step-by-step Solution:

1. **Understanding Even Functions**:
An even function satisfies the property \( f(x) = f(-x) \). This means that the function has symmetry about the y-axis.

**Hint**: Recall the definition of even functions and how they behave with respect to their arguments.

2. **Transforming the Integral**:
We want to evaluate \( \int_0^\pi f(\cos x) \, dx \). We can use the property of even functions to analyze \( f(\cos x) \).

3. **Using the Property of Cosine**:
Notice that \( \cos(\pi - x) = -\cos(x) \). Therefore, we can substitute \( u = \pi - x \) in the integral. When \( x = 0 \), \( u = \pi \) and when \( x = \pi \), \( u = 0 \). The differential \( dx = -du \).

Thus, we can rewrite the integral:
\[ \int_0^\pi f(\cos x) \, dx = \int_\pi^0 f(\cos(\pi - u)) (-du) = \int_0^\pi f(-\cos u) \, du \]

4. **Using the Even Function Property**:
Since \( f(x) \) is even, we have:
\[ f(-\cos u) = f(\cos u) \]
Therefore, we can rewrite the integral as:
\[ \int_0^\pi f(\cos x) \, dx = \int_0^\pi f(\cos u) \, du \]

5. **Combining the Integrals**:
Now we have:
\[ \int_0^\pi f(\cos x) \, dx = \int_0^\pi f(\cos x) \, dx \]
This means we can express the original integral in terms of half the interval:
\[ \int_0^\pi f(\cos x) \, dx = 2 \int_0^{\pi/2} f(\cos x) \, dx \]

6. **Final Result**:
Therefore, the final result for the integral \( \int_0^\pi f(\cos x) \, dx \) is:
\[ \int_0^\pi f(\cos x) \, dx = 2 \int_0^{\pi/2} f(\cos x) \, dx \]

### Conclusion:
The value of \( \int_0^\pi f(\cos x) \, dx \) is equal to \( 2 \int_0^{\pi/2} f(\cos x) \, dx \).