For any `t in R` and f be a continuous function
Let `I_(1)= int_(sin^(2)t)^(1+cos^(2) t) x f(x(2-x)) dx and I_(2)= int_(sin^(2)t)^(1+cos^(2)t) f(x(2-x)) dx` then `(I_(1))/(I_(2))` is

Options

A. 0
B. 1
C. 2
D. 3

Answer

To solve the problem, we need to evaluate the ratio \( \frac{I_1}{I_2} \) where:

\[ I_1 = \int_{\sin^2 t}^{1 + \cos^2 t} x \cdot f(x(2-x)) \, dx \]

\[ I_2 = \int_{\sin^2 t}^{1 + \cos^2 t} f(x(2-x)) \, dx \]

### Step 1: Use the property of definite integrals

We will use the property of definite integrals which states that:

\[ \int_a^b f(x) \, dx = \int_a^b f(a + b - x) \, dx \]

In our case, we can apply this property to \( I_1 \).

### Step 2: Determine the limits of integration

First, we need to find the sum of the limits of integration:

\[ \text{Sum of limits} = \sin^2 t + (1 + \cos^2 t) = \sin^2 t + 1 + \cos^2 t = 2 \]

### Step 3: Rewrite \( I_1 \)

Using the property mentioned earlier, we can rewrite \( I_1 \):

\[ I_1 = \int_{\sin^2 t}^{1 + \cos^2 t} x \cdot f(x(2-x)) \, dx = \int_{\sin^2 t}^{1 + \cos^2 t} (2 - x) \cdot f((2-x)x) \, dx \]

### Step 4: Combine the integrals

Now, we can express \( I_1 \) in terms of \( I_2 \):

\[ I_1 = \int_{\sin^2 t}^{1 + \cos^2 t} (2 \cdot f(x(2-x)) - x \cdot f(x(2-x))) \, dx \]

This can be split into two integrals:

\[ I_1 = 2 \int_{\sin^2 t}^{1 + \cos^2 t} f(x(2-x)) \, dx - \int_{\sin^2 t}^{1 + \cos^2 t} x \cdot f(x(2-x)) \, dx \]

### Step 5: Relate \( I_1 \) and \( I_2 \)

Notice that the second integral on the right side is actually \( I_1 \):

\[ I_1 = 2I_2 - I_1 \]

### Step 6: Solve for \( I_1 \)

Rearranging gives:

\[ 2I_1 = 2I_2 \]

Dividing both sides by 2:

\[ I_1 = I_2 \]

### Step 7: Find the ratio \( \frac{I_1}{I_2} \)

Now, we can find the ratio:

\[ \frac{I_1}{I_2} = \frac{I_2}{I_2} = 1 \]

Thus, the final answer is:

\[ \frac{I_1}{I_2} = 1 \]