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 problem, we need to evaluate the two statements regarding definite integrals.

### Statement 1:
We need to evaluate the integral:
\[ I_1 = \int_{0}^{\frac{\pi}{2}} x \cot x \, dx \]

### Step 1: Integration by Parts
We can use integration by parts, where we let:
- \( u = x \) → \( du = dx \)
- \( dv = \cot x \, dx \) → \( v = \log(\sin x) \)

Using the integration by parts formula:
\[ \int u \, dv = uv - \int v \, du \]
we have:
\[ I_1 = x \log(\sin x) \bigg|_{0}^{\frac{\pi}{2}} - \int_{0}^{\frac{\pi}{2}} \log(\sin x) \, dx \]

### Step 2: Evaluate the Boundary Terms
Now, we evaluate the boundary terms \( x \log(\sin x) \) at \( x = \frac{\pi}{2} \) and \( x = 0 \):
- At \( x = \frac{\pi}{2} \):
\[ \frac{\pi}{2} \log(\sin(\frac{\pi}{2})) = \frac{\pi}{2} \log(1) = 0 \]
- At \( x = 0 \):
\[ \lim_{x \to 0} x \log(\sin x) \]
Using L'Hôpital's rule, we can rewrite this as:
\[ \lim_{x \to 0} x \log(x) = \lim_{x \to 0} \frac{\log(x)}{1/x} \]
Differentiating the numerator and denominator:
\[ = \lim_{x \to 0} \frac{1/x}{-1/x^2} = \lim_{x \to 0} -x = 0 \]
Thus, the boundary terms evaluate to:
\[ 0 - 0 = 0 \]

### Step 3: Substitute Back
Now substituting back into the integral:
\[ I_1 = 0 - \int_{0}^{\frac{\pi}{2}} \log(\sin x) \, dx \]
Thus, we have:
\[ I_1 = -\int_{0}^{\frac{\pi}{2}} \log(\sin x) \, dx \]

### Step 4: Using Statement 2
From Statement 2, we know:
\[ \int_{0}^{\frac{\pi}{2}} \log(\sin x) \, dx = -\frac{\pi}{2} \log 2 \]
Substituting this into our expression for \( I_1 \):
\[ I_1 = -\left(-\frac{\pi}{2} \log 2\right) = \frac{\pi}{2} \log 2 \]

### Conclusion for Statement 1
Thus, we conclude that:
\[ \int_{0}^{\frac{\pi}{2}} x \cot x \, dx = \frac{\pi}{2} \log 2 \]
This confirms that Statement 1 is true.

### Summary for Statement 2
We have:
\[ \int_{0}^{\frac{\pi}{2}} \log(\sin x) \, dx = -\frac{\pi}{2} \log 2 \]
This confirms that Statement 2 is also true.

### Final Answer
Both statements are true:
- Statement 1: True
- Statement 2: True