Let the be an integrable function defined on [0,a] if `I_(1)=int_(pi//2)^(pi//2) cos theta f(sin theta cos ^(2) theta)d theta and I_(2)=int_(0)^(pi//2)sin 2 theta f(sin theta cos ^(2) theta)d theta` then

Options

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

Explanation

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

1. **Define the integrals:**
\[ I_1 = \int_{\frac{\pi}{2}}^{\frac{\pi}{2}} \cos \theta \, f(\sin \theta \cos^2 \theta) \, d\theta \]
\[ I_2 = \int_{0}^{\frac{\pi}{2}} \sin 2\theta \, f(\sin \theta \cos^2 \theta) \, d\theta \]

2. **Set up the equation:**
We want to find the relationship between \( I_1 \) and \( I_2 \). We can start by rewriting \( I_1 - I_2 \):
\[ I_1 - I_2 = \int_{0}^{\frac{\pi}{2}} \cos \theta \, f(\sin \theta \cos^2 \theta) \, d\theta - \int_{0}^{\frac{\pi}{2}} \sin 2\theta \, f(\sin \theta \cos^2 \theta) \, d\theta \]

3. **Combine the integrals:**
Since both integrals are from \( 0 \) to \( \frac{\pi}{2} \), we can combine them:
\[ I_1 - I_2 = \int_{0}^{\frac{\pi}{2}} \left( \cos \theta - \sin 2\theta \right) f(\sin \theta \cos^2 \theta) \, d\theta \]

4. **Differentiate the expression:**
Now, we differentiate the expression inside the integral:
\[ \frac{d}{d\theta}(\sin \theta + \cos^2 \theta) = \cos \theta + 2\cos \theta(-\sin \theta) = \cos \theta - \sin \theta \]
Thus, we can express \( \cos \theta - \sin 2\theta \) in terms of the derivative:
\[ \cos \theta - \sin 2\theta = \cos \theta - 2\sin \theta \cos \theta = \cos \theta (1 - 2\sin \theta) \]

5. **Substituting back:**
Substitute back into the integral:
\[ I_1 - I_2 = \int_{0}^{\frac{\pi}{2}} \cos \theta (1 - 2\sin \theta) f(\sin \theta \cos^2 \theta) \, d\theta \]

6. **Change of variable:**
Let \( t = \sin \theta + \cos^2 \theta \). Then we need to find the limits of integration. When \( \theta = 0 \), \( t = 0 + 1 = 1 \), and when \( \theta = \frac{\pi}{2} \), \( t = 1 + 0 = 1 \). Thus, the limits remain the same:
\[ I_1 - I_2 = \int_{1}^{1} f(t) \, dt = 0 \]

7. **Conclusion:**
Since \( I_1 - I_2 = 0 \), we conclude that:
\[ I_1 = I_2 \]

### Final Answer:
Thus, \( I_1 \) is equal to \( I_2 \).