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

Options

A. `(pi)/(2)`
B. `-pi`
C. `-2`
D. 1

Answer

To solve the integral \( \int_{0}^{\frac{\pi}{2}} x \sin x \, dx \), we will use the method of integration by parts.

### Step-by-Step Solution:

1. **Identify the functions for integration by parts**:
We choose:
- \( u = x \) (first function)
- \( dv = \sin x \, dx \) (second function)

2. **Differentiate and integrate**:
- Differentiate \( u \):
\[ du = dx \]
- Integrate \( dv \):
\[ v = -\cos x \]

3. **Apply the integration by parts formula**:
The formula for integration by parts is:
\[ \int u \, dv = uv - \int v \, du \]
Substituting our values:
\[ \int x \sin x \, dx = x(-\cos x) - \int (-\cos x) \, dx \]
This simplifies to:
\[ -x \cos x + \int \cos x \, dx \]

4. **Integrate \( \cos x \)**:
The integral of \( \cos x \) is:
\[ \int \cos x \, dx = \sin x \]
Therefore, we have:
\[ -x \cos x + \sin x \]

5. **Evaluate the definite integral from 0 to \( \frac{\pi}{2} \)**:
Now we need to evaluate:
\[ \left[-x \cos x + \sin x\right]_{0}^{\frac{\pi}{2}} \]
First, we evaluate at the upper limit \( x = \frac{\pi}{2} \):
\[ -\left(\frac{\pi}{2}\right) \cos\left(\frac{\pi}{2}\right) + \sin\left(\frac{\pi}{2}\right) = -\left(\frac{\pi}{2}\right) \cdot 0 + 1 = 1 \]
Next, we evaluate at the lower limit \( x = 0 \):
\[ -0 \cdot \cos(0) + \sin(0) = 0 + 0 = 0 \]

6. **Combine the results**:
Now, we subtract the lower limit result from the upper limit result:
\[ 1 - 0 = 1 \]

### Final Answer:
Thus, the value of the integral \( \int_{0}^{\frac{\pi}{2}} x \sin x \, dx \) is \( 1 \).

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