`int_(0)^(pi)x sin^(2)x dx=`

Options

A. `(pi)/(2)`
B. `(pi)/(4)`
C. `(pi^(2))/(4)`
D. `(pi^(2))/(2)`

Answer

To solve the integral \( I = \int_{0}^{\pi} x \sin^2 x \, dx \), we can use the property of definite integrals and symmetry.

### Step 1: Define the Integral
Let \( I = \int_{0}^{\pi} x \sin^2 x \, dx \).

### Step 2: Use the Property of Definite Integrals
We can use the property of definite integrals:
\[ I = \int_{0}^{\pi} (\pi - x) \sin^2 x \, dx \]
This is because if we substitute \( x \) with \( \pi - x \), the limits remain the same.

### Step 3: Rewrite the Integral
Now, we can rewrite the integral:
\[ I = \int_{0}^{\pi} (\pi - x) \sin^2 x \, dx = \int_{0}^{\pi} \pi \sin^2 x \, dx - \int_{0}^{\pi} x \sin^2 x \, dx \]
This gives us:
\[ I = \pi \int_{0}^{\pi} \sin^2 x \, dx - I \]

### Step 4: Solve for \( I \)
Now, we can add \( I \) to both sides:
\[ 2I = \pi \int_{0}^{\pi} \sin^2 x \, dx \]
Thus,
\[ I = \frac{\pi}{2} \int_{0}^{\pi} \sin^2 x \, dx \]

### Step 5: Evaluate \( \int_{0}^{\pi} \sin^2 x \, dx \)
To evaluate \( \int_{0}^{\pi} \sin^2 x \, dx \), we can use the identity:
\[ \sin^2 x = \frac{1 - \cos(2x)}{2} \]
Thus,
\[ \int_{0}^{\pi} \sin^2 x \, dx = \int_{0}^{\pi} \frac{1 - \cos(2x)}{2} \, dx \]
This simplifies to:
\[ = \frac{1}{2} \left( \int_{0}^{\pi} 1 \, dx - \int_{0}^{\pi} \cos(2x) \, dx \right) \]

### Step 6: Calculate Each Integral
The first integral is straightforward:
\[ \int_{0}^{\pi} 1 \, dx = \pi \]
The second integral can be evaluated as:
\[ \int_{0}^{\pi} \cos(2x) \, dx = \left[ \frac{\sin(2x)}{2} \right]_{0}^{\pi} = \frac{\sin(2\pi) - \sin(0)}{2} = 0 \]
Thus,
\[ \int_{0}^{\pi} \sin^2 x \, dx = \frac{1}{2} \left( \pi - 0 \right) = \frac{\pi}{2} \]

### Step 7: Substitute Back to Find \( I \)
Now substituting back:
\[ I = \frac{\pi}{2} \cdot \frac{\pi}{2} = \frac{\pi^2}{4} \]

### Final Answer
Thus, the final value of the integral is:
\[ \int_{0}^{\pi} x \sin^2 x \, dx = \frac{\pi^2}{4} \]

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