`Ifint_0^(npi)f(cos^2x)dx=kint_0^pif(cos^2x)dx , then find the value of k`

Explanation

To solve the problem, we need to evaluate the integral \( \int_0^{n\pi} f(\cos^2 x) \, dx \) and relate it to \( \int_0^{\pi} f(\cos^2 x) \, dx \).

### Step-by-Step Solution:

1. **Understanding the periodicity of \( \cos^2 x \)**:
The function \( \cos^2 x \) is periodic with a period of \( \pi \). This means that the integral of \( f(\cos^2 x) \) over any interval of length \( \pi \) will be the same.

2. **Breaking down the integral**:
We can express the integral from \( 0 \) to \( n\pi \) as a sum of integrals over intervals of length \( \pi \):
\[ \int_0^{n\pi} f(\cos^2 x) \, dx = \int_0^{\pi} f(\cos^2 x) \, dx + \int_{\pi}^{2\pi} f(\cos^2 x) \, dx + \ldots + \int_{(n-1)\pi}^{n\pi} f(\cos^2 x) \, dx \]

3. **Using the periodicity**:
Since \( \cos^2 x \) is periodic with period \( \pi \), each of these integrals is equal:
\[ \int_{\pi}^{2\pi} f(\cos^2 x) \, dx = \int_0^{\pi} f(\cos^2 x) \, dx \]
Therefore, we have:
\[ \int_0^{n\pi} f(\cos^2 x) \, dx = n \int_0^{\pi} f(\cos^2 x) \, dx \]

4. **Relating the integrals**:
From the problem statement, we know:
\[ \int_0^{n\pi} f(\cos^2 x) \, dx = k \int_0^{\pi} f(\cos^2 x) \, dx \]
Setting the two expressions for \( \int_0^{n\pi} f(\cos^2 x) \, dx \) equal gives us:
\[ n \int_0^{\pi} f(\cos^2 x) \, dx = k \int_0^{\pi} f(\cos^2 x) \, dx \]

5. **Solving for \( k \)**:
Assuming \( \int_0^{\pi} f(\cos^2 x) \, dx \neq 0 \), we can divide both sides by \( \int_0^{\pi} f(\cos^2 x) \, dx \):
\[ k = n \]

### Final Answer:
The value of \( k \) is \( n \).