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

Options

A. `pi/2-2`
B. `pi/2+2`
C. `pi/2-1`
D. `pi/2+1`

Answer

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

### Step-by-step Solution:

1. **Identify Functions for Integration by Parts**:
We choose:
- \( u = x \) (which we will differentiate)
- \( dv = \cos x \, dx \) (which we will integrate)

2. **Differentiate and Integrate**:
Now we compute \( du \) and \( v \):
- \( du = dx \)
- \( v = \int \cos x \, dx = \sin x \)

3. **Apply the Integration by Parts Formula**:
The formula for integration by parts is:
\[ \int u \, dv = uv - \int v \, du \]
Substituting our values, we get:
\[ \int_{0}^{\frac{\pi}{2}} x \cos x \, dx = \left[ x \sin x \right]_{0}^{\frac{\pi}{2}} - \int_{0}^{\frac{\pi}{2}} \sin x \, dx \]

4. **Evaluate the First Term**:
Now we evaluate \( \left[ x \sin x \right]_{0}^{\frac{\pi}{2}} \):
- At \( x = \frac{\pi}{2} \):
\[ \frac{\pi}{2} \sin\left(\frac{\pi}{2}\right) = \frac{\pi}{2} \cdot 1 = \frac{\pi}{2} \]
- At \( x = 0 \):
\[ 0 \cdot \sin(0) = 0 \]
Thus, \( \left[ x \sin x \right]_{0}^{\frac{\pi}{2}} = \frac{\pi}{2} - 0 = \frac{\pi}{2} \).

5. **Evaluate the Integral of \( \sin x \)**:
Next, we compute \( \int_{0}^{\frac{\pi}{2}} \sin x \, dx \):
\[ \int \sin x \, dx = -\cos x \]
Therefore,
\[ \int_{0}^{\frac{\pi}{2}} \sin x \, dx = \left[ -\cos x \right]_{0}^{\frac{\pi}{2}} = -\cos\left(\frac{\pi}{2}\right) - (-\cos(0)) = 0 - (-1) = 1 \]

6. **Combine the Results**:
Now substituting back into our integration by parts formula:
\[ \int_{0}^{\frac{\pi}{2}} x \cos x \, dx = \frac{\pi}{2} - 1 \]

### Final Answer:
Thus, the value of the integral is:
\[ \int_{0}^{\frac{\pi}{2}} x \cos x \, dx = \frac{\pi}{2} - 1 \]