Let `f` be an odd function then `int_(-1)^(1) (|x| +f(x) cos x) dx` is equal to

Options

A. `0`
B. `1`
C. `2`
D. none of these

Answer

To solve the integral \( \int_{-1}^{1} (|x| + f(x) \cos x) \, dx \) where \( f \) is an odd function, we can break it down into two parts: the integral of \( |x| \) and the integral of \( f(x) \cos x \).

### Step-by-step Solution:

1. **Understanding the properties of the functions:**
- Since \( f(x) \) is an odd function, it satisfies the property \( f(-x) = -f(x) \).
- The cosine function, \( \cos x \), is an even function, meaning \( \cos(-x) = \cos(x) \).
- The product of an odd function and an even function is an odd function. Therefore, \( f(x) \cos x \) is odd.

2. **Splitting the integral:**
\[ \int_{-1}^{1} (|x| + f(x) \cos x) \, dx = \int_{-1}^{1} |x| \, dx + \int_{-1}^{1} f(x) \cos x \, dx \]

3. **Evaluating the integral of \( f(x) \cos x \):**
- Since \( f(x) \cos x \) is an odd function, the integral over a symmetric interval around zero is zero:
\[ \int_{-1}^{1} f(x) \cos x \, dx = 0 \]

4. **Evaluating the integral of \( |x| \):**
- The function \( |x| \) can be split into two parts:
\[ |x| = \begin{cases} -x & \text{for } x < 0 \\ x & \text{for } x \geq 0 \end{cases} \]
- Thus, we can write:
\[ \int_{-1}^{1} |x| \, dx = \int_{-1}^{0} -x \, dx + \int_{0}^{1} x \, dx \]

5. **Calculating the integrals:**
- For the first integral:
\[ \int_{-1}^{0} -x \, dx = \left[-\frac{x^2}{2}\right]_{-1}^{0} = \left[0 - \frac{(-1)^2}{2}\right] = -\frac{1}{2} \]
- For the second integral:
\[ \int_{0}^{1} x \, dx = \left[\frac{x^2}{2}\right]_{0}^{1} = \left[\frac{1^2}{2} - 0\right] = \frac{1}{2} \]

6. **Combining the results:**
\[ \int_{-1}^{1} |x| \, dx = -\frac{1}{2} + \frac{1}{2} = 1 \]

7. **Final result:**
\[ \int_{-1}^{1} (|x| + f(x) \cos x) \, dx = 1 + 0 = 1 \]

### Conclusion:
The value of the integral \( \int_{-1}^{1} (|x| + f(x) \cos x) \, dx \) is \( 1 \).