Which of the following definite integral (s) vanishes ?

Options

A. `int_(0)^(pi//2) In(cot x)dx`
B. `int_(0)^(pi2) sin ^(3) x dx`
C. `int_(1//e)^(e)(dx)/(x(Inx)^(1//3))`
D. `int_(0)^(pi) sqrt((1+ cos 2x)/(2))dx`

Answer

To determine which of the following definite integrals vanish, we will evaluate each option step by step.

### Given Options:
1. \( I_1 = \int_0^{\frac{\pi}{2}} \ln(\cot x) \, dx \)
2. \( I_2 = \int_0^{\frac{\pi}{2}} \sin^3 x \, dx \)
3. \( I_3 = \int_1^e \frac{dx}{x \ln(x)^{\frac{1}{3}}} \)
4. \( I_4 = \int_0^{\pi} \sqrt{\frac{1 + \cos(2x)}{2}} \, dx \)

### Step 1: Evaluate \( I_1 = \int_0^{\frac{\pi}{2}} \ln(\cot x) \, dx \)

1. **Using the property of definite integrals:**
\[ I_1 = \int_0^{\frac{\pi}{2}} \ln(\cot x) \, dx = \int_0^{\frac{\pi}{2}} \ln(\tan(\frac{\pi}{2} - x)) \, dx \]
This means:
\[ I_1 = \int_0^{\frac{\pi}{2}} \ln(\tan x) \, dx \]
Let \( I = \int_0^{\frac{\pi}{2}} \ln(\tan x) \, dx \).

2. **Adding both expressions:**
\[ 2I = \int_0^{\frac{\pi}{2}} \ln(\tan x) + \ln(\cot x) \, dx = \int_0^{\frac{\pi}{2}} \ln(1) \, dx = 0 \]
Therefore, \( I = 0 \).

### Conclusion for \( I_1 \):
\[ I_1 \text{ vanishes.} \]

### Step 2: Evaluate \( I_2 = \int_0^{\frac{\pi}{2}} \sin^3 x \, dx \)

1. **Using the identity:**
\[ \sin^3 x = \sin x (1 - \cos^2 x) \]
Thus,
\[ I_2 = \int_0^{\frac{\pi}{2}} \sin x \, dx - \int_0^{\frac{\pi}{2}} \sin x \cos^2 x \, dx \]

2. **Calculating \( \int_0^{\frac{\pi}{2}} \sin x \, dx = 1 \)** and using substitution for \( \int_0^{\frac{\pi}{2}} \sin x \cos^2 x \, dx \):
\[ \text{Let } u = \cos x \Rightarrow du = -\sin x \, dx \]
Changing limits from \( 1 \) to \( 0 \):
\[ \int_0^{\frac{\pi}{2}} \sin x \cos^2 x \, dx = \int_1^0 u^2 (-du) = \int_0^1 u^2 \, du = \frac{1}{3} \]
Thus,
\[ I_2 = 1 - \frac{1}{3} = \frac{2}{3} \neq 0 \]

### Conclusion for \( I_2 \):
\[ I_2 \text{ does not vanish.} \]

### Step 3: Evaluate \( I_3 = \int_1^e \frac{dx}{x \ln(x)^{\frac{1}{3}}} \)

1. **Using substitution:**
Let \( t = \ln x \Rightarrow dt = \frac{1}{x} dx \).
Changing limits:
- When \( x = 1 \), \( t = 0 \)
- When \( x = e \), \( t = 1 \)

Thus,
\[ I_3 = \int_0^1 \frac{dt}{t^{\frac{1}{3}}} = 3t^{\frac{2}{3}} \bigg|_0^1 = 3(1 - 0) = 3 \neq 0 \]

### Conclusion for \( I_3 \):
\[ I_3 \text{ does not vanish.} \]

### Step 4: Evaluate \( I_4 = \int_0^{\pi} \sqrt{\frac{1 + \cos(2x)}{2}} \, dx \)

1. **Using the identity:**
\[ \sqrt{\frac{1 + \cos(2x)}{2}} = \sqrt{\cos^2 x} = |\cos x| \]
Since \( \cos x \) is non-negative from \( 0 \) to \( \frac{\pi}{2} \) and non-positive from \( \frac{\pi}{2} \) to \( \pi \):
\[ I_4 = \int_0^{\frac{\pi}{2}} \cos x \, dx + \int_{\frac{\pi}{2}}^{\pi} -\cos x \, dx \]
Both integrals yield:
\[ = \sin x \bigg|_0^{\frac{\pi}{2}} + (-\sin x) \bigg|_{\frac{\pi}{2}}^{\pi} = 1 + 1 = 2 \neq 0 \]

### Conclusion for \( I_4 \):
\[ I_4 \text{ does not vanish.} \]

### Final Conclusion:
The definite integrals that vanish are:
- \( I_1 \) and \( I_3 \)