Statement-1: `int_(0)^(pi//2) x cot x dx=(pi)/(2)log2`
Statement-2: `int_(0)^(pi//2) log sin x dx=-(pi)/(2)log2`

Options

A. Statement-1 is true, Statement-2 is True,Statement-2 is a correct explanation for Statement-1.
B. Statement-1 is True, Statement-2 is not a correct explanation for Statement-1.
C. Statement-1 is True, Statement-2 is False.
D. Statement-1 is False, Statement-2 is True.

Explanation

To solve the given problem, we will analyze both statements step by step.

### Step 1: Evaluate Statement 2
We start with the second statement:
\[ \int_{0}^{\frac{\pi}{2}} \log(\sin x) \, dx = -\frac{\pi}{2} \log 2 \]

This result is a well-known integral. To understand why this is true, we can use the property of definite integrals. We can also consider the integral of \(\log(\cos x)\):
\[ \int_{0}^{\frac{\pi}{2}} \log(\cos x) \, dx \]
Using the substitution \(x = \frac{\pi}{2} - t\), we find that:
\[ \int_{0}^{\frac{\pi}{2}} \log(\cos x) \, dx = \int_{0}^{\frac{\pi}{2}} \log(\sin t) \, dt \]
This means:
\[ \int_{0}^{\frac{\pi}{2}} \log(\sin x) \, dx + \int_{0}^{\frac{\pi}{2}} \log(\cos x) \, dx = \int_{0}^{\frac{\pi}{2}} \log(\sin x \cos x) \, dx \]
Using the identity \(\sin x \cos x = \frac{1}{2} \sin(2x)\), we have:
\[ \int_{0}^{\frac{\pi}{2}} \log(\sin x \cos x) \, dx = \int_{0}^{\frac{\pi}{2}} \log\left(\frac{1}{2} \sin(2x)\right) \, dx \]
This simplifies to:
\[ \int_{0}^{\frac{\pi}{2}} \log\left(\frac{1}{2}\right) \, dx + \int_{0}^{\frac{\pi}{2}} \log(\sin(2x)) \, dx \]
The first integral evaluates to \(-\frac{\pi}{2} \log(2)\), and the second integral can be evaluated using the substitution \(u = 2x\), yielding:
\[ \frac{1}{2} \int_{0}^{\pi} \log(\sin u) \, du = \frac{1}{2} \cdot -\frac{\pi}{2} \log(2) = -\frac{\pi}{4} \log(2) \]
Combining these results, we find:
\[ \int_{0}^{\frac{\pi}{2}} \log(\sin x) \, dx + \int_{0}^{\frac{\pi}{2}} \log(\cos x) \, dx = -\frac{\pi}{2} \log(2) \]
Since both integrals are equal, we conclude:
\[ 2 \int_{0}^{\frac{\pi}{2}} \log(\sin x) \, dx = -\frac{\pi}{2} \log(2) \implies \int_{0}^{\frac{\pi}{2}} \log(\sin x) \, dx = -\frac{\pi}{2} \log(2) \]

### Step 2: Evaluate Statement 1
Now we evaluate the first statement:
\[ \int_{0}^{\frac{\pi}{2}} x \cot x \, dx \]
We will use integration by parts. Let:
- \(u = x\) \(\Rightarrow du = dx\)
- \(dv = \cot x \, dx\) \(\Rightarrow v = \log(\sin x)\)

Using integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
Thus,
\[ \int_{0}^{\frac{\pi}{2}} x \cot x \, dx = \left[ x \log(\sin x) \right]_{0}^{\frac{\pi}{2}} - \int_{0}^{\frac{\pi}{2}} \log(\sin x) \, dx \]
Evaluating the boundary term:
At \(x = \frac{\pi}{2}\), \(\log(\sin(\frac{\pi}{2})) = \log(1) = 0\).
At \(x = 0\), \(x \log(\sin x) \to 0\) as \(x \to 0\).

Thus, the boundary term evaluates to \(0\):
\[ \left[ x \log(\sin x) \right]_{0}^{\frac{\pi}{2}} = 0 - 0 = 0 \]
Now substituting the value of \(\int_{0}^{\frac{\pi}{2}} \log(\sin x) \, dx\):
\[ \int_{0}^{\frac{\pi}{2}} x \cot x \, dx = 0 - \left(-\frac{\pi}{2} \log(2)\right) = \frac{\pi}{2} \log(2) \]

### Conclusion
Both statements are correct:
1. \(\int_{0}^{\frac{\pi}{2}} x \cot x \, dx = \frac{\pi}{2} \log(2)\)
2. \(\int_{0}^{\frac{\pi}{2}} \log(\sin x) \, dx = -\frac{\pi}{2} \log(2)\)