For `x epsilonR`, and a continuous function `f` let `I_(1)=int_(sin^(2)t)^(1+cos^(2)t)xf{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. `-1`
B. `1`
C. `2`
D. `3`

Explanation

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

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

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

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

We will use the property of definite integrals which states:

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

In our case, \( a = \sin^2 t \) and \( b = 1 + \cos^2 t \). First, we calculate \( a + b \):

\[ a + b = \sin^2 t + (1 + \cos^2 t) = \sin^2 t + 1 + \cos^2 t = 2 \]

### Step 2: Apply the property to \( I_1 \)

Now, we apply this property to \( I_1 \):

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

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

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

### Step 3: Combine the original and transformed integrals

Now we can add the original \( I_1 \) and the transformed version:

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

### Step 4: Simplify the expression

This can be simplified as:

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

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

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

\[ 2I_1 = 2I_2 \]

Dividing both sides by 2 gives:

\[ I_1 = I_2 \]

### Step 6: Calculate the ratio

Now, we can find the ratio:

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

### Final Answer

Thus, the final answer is:

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