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

Options

A. `I_(1)=I_(2)`
B. `I_(1)=2I_(2)`
C. `I_(1)=5I_(2)`
D. none of these

Answer

To solve the problem, we need to find the relationship between the integrals \( I_1 \) and \( I_2 \).

### Step 1: Define the integrals
We have:
\[ I_1 = \int_{0}^{3\pi} f(\cos^2 x) \, dx \]
\[ I_2 = \int_{0}^{\pi} f(\cos^2 x) \, dx \]

### Step 2: Break down \( I_1 \)
We can break down \( I_1 \) into three intervals:
\[ I_1 = \int_{0}^{\pi} f(\cos^2 x) \, dx + \int_{\pi}^{2\pi} f(\cos^2 x) \, dx + \int_{2\pi}^{3\pi} f(\cos^2 x) \, dx \]

### Step 3: Evaluate \( \int_{\pi}^{2\pi} f(\cos^2 x) \, dx \)
Using the substitution \( x = \pi + t \) where \( t \) goes from \( 0 \) to \( \pi \):
\[ \int_{\pi}^{2\pi} f(\cos^2 x) \, dx = \int_{0}^{\pi} f(\cos^2(\pi + t)) \, dt \]
Since \( \cos(\pi + t) = -\cos t \), we have \( \cos^2(\pi + t) = \cos^2 t \):
\[ \int_{\pi}^{2\pi} f(\cos^2 x) \, dx = \int_{0}^{\pi} f(\cos^2 t) \, dt = I_2 \]

### Step 4: Evaluate \( \int_{2\pi}^{3\pi} f(\cos^2 x) \, dx \)
Using the substitution \( x = 2\pi + t \) where \( t \) goes from \( 0 \) to \( \pi \):
\[ \int_{2\pi}^{3\pi} f(\cos^2 x) \, dx = \int_{0}^{\pi} f(\cos^2(2\pi + t)) \, dt \]
Since \( \cos(2\pi + t) = \cos t \), we have \( \cos^2(2\pi + t) = \cos^2 t \):
\[ \int_{2\pi}^{3\pi} f(\cos^2 x) \, dx = \int_{0}^{\pi} f(\cos^2 t) \, dt = I_2 \]

### Step 5: Combine the results
Now we can combine the results:
\[ I_1 = I_2 + I_2 + I_2 = 3 I_2 \]

### Conclusion
Thus, we have established that:
\[ I_1 = 3 I_2 \]

### Final Answer
The relationship between the integrals is:
\[ I_1 = 3 I_2 \]