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 in the question 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: Use the property of definite integrals

We can use the property of definite integrals that states:

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

for any periodic function \( g(x) \) with period \( \pi \). In our case, \( g(x) = f(\cos^2 x) \).

### Step 2: Apply the property to our integral

Since \( \cos^2 x \) is a periodic function with period \( \pi \), we can apply the property:

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

### Step 3: Set the equation

Now we can set the left-hand side equal to the right-hand side, as given in the problem:

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

### Step 4: Solve for \( k \)

To find \( k \), we can divide both sides of the equation by \( \int_{0}^{\pi} f(\cos^2 x) \, dx \) (assuming it is not zero):

\[ k = n \]

### Conclusion

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

\[ \boxed{n} \]

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