If `I_(1)= int_(0)^(pi) x f {sin^(3) x +cos^(2)x} dx and I_(2)= pi int_(0)^(pi//2) f (sin^(3)x + cos^(2) x}dx` then

Options

A. `I_(1)= I_(2)`
B. `I_(1) + I_(2)= 0`
C. `I_(1)= 2I_(2)`
D. none

Answer

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

### Step-by-Step Solution

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

2. **Use the Property of Integrals**:
We will use the property of integrals:
\[ \int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx \]
Applying this to \( I_1 \):
\[ I_1 = \int_0^{\pi} (\pi - x) f(\sin^3(\pi - x) + \cos^2(\pi - x)) \, dx \]

3. **Simplify the Function**:
We know that:
\[ \sin(\pi - x) = \sin x \quad \text{and} \quad \cos(\pi - x) = -\cos x \]
Therefore:
\[ \cos^2(\pi - x) = \cos^2 x \]
Thus:
\[ I_1 = \int_0^{\pi} (\pi - x) f(\sin^3 x + \cos^2 x) \, dx \]

4. **Split the Integral**:
We can split \( I_1 \) into two parts:
\[ I_1 = \pi \int_0^{\pi} f(\sin^3 x + \cos^2 x) \, dx - \int_0^{\pi} x f(\sin^3 x + \cos^2 x) \, dx \]
The second integral is \( I_1 \) itself, so we can write:
\[ 2I_1 = \pi \int_0^{\pi} f(\sin^3 x + \cos^2 x) \, dx \]

5. **Change the Limits**:
Now, we use the property:
\[ \int_0^a f(x) \, dx = 2 \int_0^{\frac{a}{2}} f(x) \, dx \]
Applying this to the integral:
\[ \int_0^{\pi} f(\sin^3 x + \cos^2 x) \, dx = 2 \int_0^{\frac{\pi}{2}} f(\sin^3 x + \cos^2 x) \, dx \]
Therefore:
\[ 2I_1 = \pi \cdot 2 \int_0^{\frac{\pi}{2}} f(\sin^3 x + \cos^2 x) \, dx = 2\pi \int_0^{\frac{\pi}{2}} f(\sin^3 x + \cos^2 x) \, dx \]

6. **Relate \( I_1 \) and \( I_2 \)**:
Since \( I_2 = \pi \int_0^{\frac{\pi}{2}} f(\sin^3 x + \cos^2 x) \, dx \), we can write:
\[ 2I_1 = 2I_2 \]
Thus, we conclude:
\[ I_1 = I_2 \]

### Final Result
The relationship between \( I_1 \) and \( I_2 \) is:
\[ I_1 = I_2 \]