If `I_(1)= int_(0)^(3pi) f (cos^(2) x)dx and I_(2)= int_(0)^(pi) f (cos^(2)x) dx`, then

Options

A. `I_(1)= I_(2)`
B. `I_(1)= 2I_(2)`
C. `I_(1)= 5I_(2)`
D. `I_(1)= 3I_(2)`

Answer

To solve the problem, we need to analyze the integrals \( I_1 \) and \( I_2 \) given in the question.

1. **Define the integrals**:
\[ I_1 = \int_0^{3\pi} f(\cos^2 x) \, dx \]
\[ I_2 = \int_0^{\pi} f(\cos^2 x) \, dx \]

2. **Identify the periodicity of the function**:
The function \( \cos^2 x \) has a period of \( \pi \). This means that:
\[ \cos^2(x + \pi) = \cos^2 x \]

3. **Apply the periodicity property**:
We can use the property of integrals for periodic functions:
\[ \int_0^{nT} f(x) \, dx = n \int_0^T f(x) \, dx \]
where \( T \) is the period of the function. In our case, \( T = \pi \).

4. **Calculate \( I_1 \)**:
Since \( I_1 \) spans from \( 0 \) to \( 3\pi \), we can break it down using the periodicity:
\[ I_1 = \int_0^{3\pi} f(\cos^2 x) \, dx = 3 \int_0^{\pi} f(\cos^2 x) \, dx = 3I_2 \]

5. **Final relationship**:
From the above calculation, we conclude that:
\[ I_1 = 3I_2 \]

### Summary of Steps:
1. Define the integrals \( I_1 \) and \( I_2 \).
2. Identify the periodicity of \( \cos^2 x \).
3. Apply the periodicity property to \( I_1 \).
4. Calculate \( I_1 \) in terms of \( I_2 \).
5. State the final relationship \( I_1 = 3I_2 \).