`int_(0)^(pi//2) x sin x dx ` is equal to

Options

A. 0
B. 1
C. `-1`
D. 2

Explanation

To solve the integral \( I = \int_{0}^{\frac{\pi}{2}} x \sin x \, dx \), we will use the method of integration by parts. The formula for integration by parts is given by:

\[ \int u \, dv = uv - \int v \, du \]

### Step 1: Choose \( u \) and \( dv \)
Let:
- \( u = x \) (which means \( du = dx \))
- \( dv = \sin x \, dx \) (which means \( v = -\cos x \))

### Step 2: Apply the integration by parts formula
Using the integration by parts formula, we have:

\[ I = \int_{0}^{\frac{\pi}{2}} x \sin x \, dx = \left[ -x \cos x \right]_{0}^{\frac{\pi}{2}} - \int_{0}^{\frac{\pi}{2}} (-\cos x) \, dx \]

### Step 3: Evaluate the boundary term
Now we evaluate the boundary term \( \left[ -x \cos x \right]_{0}^{\frac{\pi}{2}} \):

\[ \left[ -x \cos x \right]_{0}^{\frac{\pi}{2}} = -\left( \frac{\pi}{2} \cos\left(\frac{\pi}{2}\right) - 0 \cdot \cos(0) \right) = -\left( \frac{\pi}{2} \cdot 0 - 0 \cdot 1 \right) = 0 \]

### Step 4: Evaluate the integral
Now we need to evaluate the integral \( \int_{0}^{\frac{\pi}{2}} \cos x \, dx \):

\[ \int_{0}^{\frac{\pi}{2}} \cos x \, dx = \left[ \sin x \right]_{0}^{\frac{\pi}{2}} = \sin\left(\frac{\pi}{2}\right) - \sin(0) = 1 - 0 = 1 \]

### Step 5: Combine results
Putting it all together, we have:

\[ I = 0 + 1 = 1 \]

Thus, the value of the integral \( \int_{0}^{\frac{\pi}{2}} x \sin x \, dx \) is equal to \( 1 \).

### Final Answer:
\[ \int_{0}^{\frac{\pi}{2}} x \sin x \, dx = 1 \]