Consider the definite integrals `A=int_(0)^(pi)sinx cosx^(2)xdx` and `B=int_(0)^((pi)/(2))sinx cos^(2)xdx`. Then,

Options

A. `A=2B`
B. `A=piB`
C. `A=(pi)/(2)B`
D. `B=2A`

Answer

To solve the problem, we need to evaluate the definite integrals \( A \) and \( B \) and establish a relationship between them.

1. **Define the integrals**:
\[ A = \int_{0}^{\pi} \sin x \cos^2 x \, dx \]
\[ B = \int_{0}^{\frac{\pi}{2}} \sin x \cos^2 x \, dx \]

2. **Use the property of definite integrals**:
We can use the property of integrals that states:
\[ \int_{0}^{2A} f(x) \, dx = \int_{0}^{A} f(x) \, dx + \int_{0}^{A} f(2A - x) \, dx \]
Here, we will apply it to \( A \) with \( A = \pi/2 \):
\[ A = \int_{0}^{\pi} \sin x \cos^2 x \, dx = \int_{0}^{\frac{\pi}{2}} \sin x \cos^2 x \, dx + \int_{0}^{\frac{\pi}{2}} \sin(\pi - x) \cos^2(\pi - x) \, dx \]

3. **Evaluate the second integral**:
For the second integral, we use the identities:
\[ \sin(\pi - x) = \sin x \quad \text{and} \quad \cos(\pi - x) = -\cos x \]
Since we have \( \cos^2(\pi - x) = \cos^2 x \), we can write:
\[ \int_{0}^{\frac{\pi}{2}} \sin(\pi - x) \cos^2(\pi - x) \, dx = \int_{0}^{\frac{\pi}{2}} \sin x \cos^2 x \, dx = B \]

4. **Combine the results**:
Therefore, we have:
\[ A = B + B = 2B \]

5. **Conclusion**:
Thus, we conclude that:
\[ A = 2B \]

### Final Answer:
The relationship between the integrals is \( A = 2B \).