The value of `int_(-pi)^(pi) sinx f(cosx)dx` is

Options

A. `pi`
B. `2pi`
C. 2f(1)
D. none of these

Answer

To solve the integral \( I = \int_{-\pi}^{\pi} \sin x \cdot f(\cos x) \, dx \), we will analyze the function \( g(x) = \sin x \cdot f(\cos x) \) to determine if it is an odd or even function.

### Step 1: Define the function
Let:
\[ g(x) = \sin x \cdot f(\cos x) \]

### Step 2: Check if \( g(x) \) is odd or even
To determine whether \( g(x) \) is odd or even, we will evaluate \( g(-x) \):
\[ g(-x) = \sin(-x) \cdot f(\cos(-x)) \]
Using the properties of sine and cosine:
\[ g(-x) = -\sin x \cdot f(\cos x) \]

### Step 3: Relate \( g(-x) \) to \( g(x) \)
From the expression we derived:
\[ g(-x) = -\sin x \cdot f(\cos x) = -g(x) \]
This shows that \( g(x) \) is an odd function.

### Step 4: Apply the property of integrals of odd functions
Since \( g(x) \) is an odd function, we can use the property of definite integrals:
\[ \int_{-a}^{a} g(x) \, dx = 0 \quad \text{if } g(x) \text{ is odd} \]
Thus, we have:
\[ I = \int_{-\pi}^{\pi} g(x) \, dx = 0 \]

### Conclusion
The value of the integral \( \int_{-\pi}^{\pi} \sin x \cdot f(\cos x) \, dx \) is:
\[ \boxed{0} \]