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

Options

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

Answer

To solve the integral \( I = \int_0^{\frac{\pi}{2}} x \cot x \, dx \), we will use integration by parts. The integration by parts formula is given by:

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

### Step 1: Choose \( u \) and \( dv \)

Let:
- \( u = x \) (which means \( du = dx \))
- \( dv = \cot x \, dx \) (which means we need to find \( v \))

### Step 2: Find \( v \)

To find \( v \), we need to integrate \( dv \):
\[ v = \int \cot x \, dx = \log(\sin x) + C \]

### Step 3: Apply Integration by Parts

Now we apply the integration by parts formula:
\[ I = uv \bigg|_0^{\frac{\pi}{2}} - \int v \, du \]
Substituting \( u \), \( v \), and their derivatives:
\[ I = \left[ x \log(\sin x) \right]_0^{\frac{\pi}{2}} - \int_0^{\frac{\pi}{2}} \log(\sin x) \, dx \]

### Step 4: Evaluate the Boundary Term

Now we evaluate the boundary term:
\[ \left[ x \log(\sin x) \right]_0^{\frac{\pi}{2}} = \left( \frac{\pi}{2} \log(\sin(\frac{\pi}{2})) \right) - \left( 0 \cdot \log(\sin(0)) \right) \]
Since \( \sin(\frac{\pi}{2}) = 1 \) and \( \log(1) = 0 \):
\[ = \frac{\pi}{2} \cdot 0 - 0 = 0 \]

### Step 5: Substitute Back into the Integral

Now substituting back into our equation for \( I \):
\[ I = 0 - \int_0^{\frac{\pi}{2}} \log(\sin x) \, dx \]
Thus,
\[ I = - \int_0^{\frac{\pi}{2}} \log(\sin x) \, dx \]

### Step 6: Use the Known Result

There is a known result for the integral:
\[ \int_0^{\frac{\pi}{2}} \log(\sin x) \, dx = -\frac{\pi}{2} \log(2) \]
Substituting this into our equation for \( I \):
\[ I = -\left(-\frac{\pi}{2} \log(2)\right) = \frac{\pi}{2} \log(2) \]

### Final Answer

Thus, the value of the integral is:
\[ \int_0^{\frac{\pi}{2}} x \cot x \, dx = \frac{\pi}{2} \log(2) \]