If `int_(0)^(npi) f(cos^(2)x)dx=k int_(0)^(pi) f(cos^(2)x)dx`, then the value of k, is

Options

A. 1
B. n
C. `n//2`
D. none of these

Answer

To solve the problem, we need to evaluate the integral given and find the value of \( k \) such that:

\[ \int_{0}^{n\pi} f(\cos^2 x) \, dx = k \int_{0}^{\pi} f(\cos^2 x) \, dx \]

### Step 1: Identify the Period of the Function
The function \( \cos^2 x \) has a period of \( \pi \). This means that the integral over any interval of length \( \pi \) will yield the same result.

**Hint:** Remember that the period of a function helps in breaking down the integral over larger intervals.

### Step 2: Break Down the Integral
We can express the integral from \( 0 \) to \( n\pi \) as a sum of integrals over intervals of length \( \pi \):

\[ \int_{0}^{n\pi} f(\cos^2 x) \, dx = \int_{0}^{\pi} f(\cos^2 x) \, dx + \int_{\pi}^{2\pi} f(\cos^2 x) \, dx + \ldots + \int_{(n-1)\pi}^{n\pi} f(\cos^2 x) \, dx \]

Since \( \cos^2 x \) is periodic with period \( \pi \), each of these integrals is equal to \( \int_{0}^{\pi} f(\cos^2 x) \, dx \).

**Hint:** Use the periodicity of the function to simplify the integral over multiple periods.

### Step 3: Count the Number of Intervals
There are \( n \) intervals of length \( \pi \) in the range from \( 0 \) to \( n\pi \). Therefore, we can write:

\[ \int_{0}^{n\pi} f(\cos^2 x) \, dx = n \int_{0}^{\pi} f(\cos^2 x) \, dx \]

**Hint:** The number of complete periods in the interval gives you a multiplier for the integral over one period.

### Step 4: Set Up the Equation
Now, we can set this equal to the expression given in the problem:

\[ n \int_{0}^{\pi} f(\cos^2 x) \, dx = k \int_{0}^{\pi} f(\cos^2 x) \, dx \]

**Hint:** Equating both sides allows you to isolate \( k \).

### Step 5: Solve for \( k \)
Assuming \( \int_{0}^{\pi} f(\cos^2 x) \, dx \) is not zero, we can divide both sides by \( \int_{0}^{\pi} f(\cos^2 x) \, dx \):

\[ k = n \]

### Conclusion
Thus, the value of \( k \) is:

\[ \boxed{n} \]