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 find the relationship between the definite integrals \( A \) and \( B \) defined as follows:

\[ A = \int_{0}^{\pi} \sin x \cos^2(x) \, dx \]
\[ B = \int_{0}^{\frac{\pi}{2}} \sin x \cos^2(x) \, dx \]

### Step 1: Rewrite Integral \( A \)

We can use the property of definite integrals that states:

\[ \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx \]

In our case, we can set \( a = 0 \) and \( b = \pi \). Thus, we can rewrite \( A \):

\[ A = \int_{0}^{\pi} \sin x \cos^2(x) \, dx = \int_{0}^{\pi} \sin(\pi - x) \cos^2(\pi - x) \, dx \]

### Step 2: Simplify the Integral

Using the identities \( \sin(\pi - x) = \sin x \) and \( \cos(\pi - x) = -\cos x \), we can rewrite \( \cos^2(\pi - x) \) as:

\[ \cos^2(\pi - x) = \cos^2(x) \]

Thus, we have:

\[ A = \int_{0}^{\pi} \sin x \cos^2(x) \, dx \]

### Step 3: Split the Integral

Now we can split the integral \( A \) into two parts:

\[ A = \int_{0}^{\frac{\pi}{2}} \sin x \cos^2(x) \, dx + \int_{\frac{\pi}{2}}^{\pi} \sin x \cos^2(x) \, dx \]

### Step 4: Change of Variable in the Second Integral

For the second integral, we can make a substitution. Let \( x = \pi - t \), then \( dx = -dt \). Changing the limits accordingly, when \( x = \frac{\pi}{2} \), \( t = \frac{\pi}{2} \) and when \( x = \pi \), \( t = 0 \):

\[ \int_{\frac{\pi}{2}}^{\pi} \sin x \cos^2(x) \, dx = \int_{\frac{\pi}{2}}^{0} \sin(\pi - t) \cos^2(\pi - t)(-dt) = \int_{0}^{\frac{\pi}{2}} \sin t \cos^2 t \, dt \]

Using the identities again, we find:

\[ \int_{\frac{\pi}{2}}^{\pi} \sin x \cos^2(x) \, dx = \int_{0}^{\frac{\pi}{2}} \sin t \cos^2 t \, dt = B \]

### Step 5: Combine the Results

Now we can combine the results:

\[ A = B + B = 2B \]

### Step 6: Final Relation

Thus, we have:

\[ A = 2B \]

### Conclusion

The relationship between \( A \) and \( B \) is:

\[ A = 2B \]