`int_(0)^(pi) x sin x. cos^(2) x dx`

Answer

To solve the integral \( I = \int_{0}^{\pi} x \sin x \cos^2 x \, dx \), we will use a property of definite integrals. Let's go through the steps:

### Step 1: Set up the integral
We start with the integral:
\[ I = \int_{0}^{\pi} x \sin x \cos^2 x \, dx \]

### Step 2: Use the property of definite integrals
We can use the property of definite integrals which states that:
\[ \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx \]
In our case, \( a = 0 \) and \( b = \pi \), so we have:
\[ I = \int_{0}^{\pi} (\pi - x) \sin(\pi - x) \cos^2(\pi - x) \, dx \]

### Step 3: Simplify the integral
Using the trigonometric identities:
- \( \sin(\pi - x) = \sin x \)
- \( \cos(\pi - x) = -\cos x \) (but since we have \( \cos^2(\pi - x) \), it will remain \( \cos^2 x \))

Thus, we can rewrite the integral:
\[ I = \int_{0}^{\pi} (\pi - x) \sin x \cos^2 x \, dx \]

### Step 4: Combine the two integrals
Now we have two expressions for \( I \):
1. \( I = \int_{0}^{\pi} x \sin x \cos^2 x \, dx \)
2. \( I = \int_{0}^{\pi} (\pi - x) \sin x \cos^2 x \, dx \)

Adding these two equations gives:
\[ 2I = \int_{0}^{\pi} \left( x + (\pi - x) \right) \sin x \cos^2 x \, dx \]
This simplifies to:
\[ 2I = \int_{0}^{\pi} \pi \sin x \cos^2 x \, dx \]

### Step 5: Factor out constants
We can factor out the constant \( \pi \):
\[ 2I = \pi \int_{0}^{\pi} \sin x \cos^2 x \, dx \]

### Step 6: Solve the integral
Now we need to evaluate the integral \( \int_{0}^{\pi} \sin x \cos^2 x \, dx \). We can use the substitution \( u = \cos x \), which gives \( du = -\sin x \, dx \). The limits change as follows:
- When \( x = 0 \), \( u = 1 \)
- When \( x = \pi \), \( u = -1 \)

Thus, the integral becomes:
\[ \int_{0}^{\pi} \sin x \cos^2 x \, dx = -\int_{1}^{-1} u^2 \, du = \int_{-1}^{1} u^2 \, du \]

### Step 7: Evaluate the new integral
Now we evaluate:
\[ \int_{-1}^{1} u^2 \, du = \left[ \frac{u^3}{3} \right]_{-1}^{1} = \frac{1^3}{3} - \frac{(-1)^3}{3} = \frac{1}{3} - \left(-\frac{1}{3}\right) = \frac{2}{3} \]

### Step 8: Substitute back to find \( I \)
Substituting back, we have:
\[ 2I = \pi \cdot \frac{2}{3} \]
Thus,
\[ I = \frac{\pi}{3} \]

### Final Answer
The value of the integral is:
\[ \int_{0}^{\pi} x \sin x \cos^2 x \, dx = \frac{\pi}{3} \]