Evaluate
(i) `int_(0)^(pi//2)(d x)/(1+sqrt(tan x))`
(ii) `int_(0)^(pi//2) log (tan x ) d x`
(iii) `int_(0)^(pi//4) log (1+tan x ) d x`
(iv) `int_(0)^(pi//2)(sin x- cos x)/(1+ sin x cos x)d x`

Answer

Let's evaluate the given integrals step by step.

### Part (i): Evaluate \( I = \int_{0}^{\frac{\pi}{2}} \frac{dx}{1 + \sqrt{\tan x}} \)

1. **Using the property of definite integrals:**
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{dx}{1 + \sqrt{\tan x}} = \int_{0}^{\frac{\pi}{2}} \frac{dx}{1 + \sqrt{\cot x}} \]
This is because \( \tan\left(\frac{\pi}{2} - x\right) = \cot x \).

2. **Setting up the equation:**
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{dx}{1 + \sqrt{\tan x}} + \int_{0}^{\frac{\pi}{2}} \frac{dx}{1 + \sqrt{\cot x}} \]
Let \( J = \int_{0}^{\frac{\pi}{2}} \frac{dx}{1 + \sqrt{\cot x}} \).

3. **Adding the two integrals:**
\[ I + J = \int_{0}^{\frac{\pi}{2}} \left( \frac{1}{1 + \sqrt{\tan x}} + \frac{1}{1 + \sqrt{\cot x}} \right) dx \]

4. **Simplifying the integrand:**
\[ \frac{1}{1 + \sqrt{\tan x}} + \frac{1}{1 + \sqrt{\cot x}} = \frac{(1 + \sqrt{\cot x}) + (1 + \sqrt{\tan x})}{(1 + \sqrt{\tan x})(1 + \sqrt{\cot x})} \]
The numerator simplifies to \( 2 + \sqrt{\tan x} + \sqrt{\cot x} \).

5. **Using the identity \( \sqrt{\tan x} + \sqrt{\cot x} = 2 \):**
\[ I + J = \int_{0}^{\frac{\pi}{2}} dx = \frac{\pi}{2} \]

6. **Since \( I = J \):**
\[ 2I = \frac{\pi}{2} \implies I = \frac{\pi}{4} \]

### Part (ii): Evaluate \( I = \int_{0}^{\frac{\pi}{2}} \log(\tan x) \, dx \)

1. **Using the property of definite integrals:**
\[ I = \int_{0}^{\frac{\pi}{2}} \log(\tan x) \, dx = \int_{0}^{\frac{\pi}{2}} \log(\cot x) \, dx \]

2. **Adding the two integrals:**
\[ I + I = \int_{0}^{\frac{\pi}{2}} \left( \log(\tan x) + \log(\cot x) \right) dx = \int_{0}^{\frac{\pi}{2}} \log(1) \, dx = 0 \]

3. **Thus, we have:**
\[ 2I = 0 \implies I = 0 \]

### Part (iii): Evaluate \( I = \int_{0}^{\frac{\pi}{4}} \log(1 + \tan x) \, dx \)

1. **Using the property of definite integrals:**
\[ I = \int_{0}^{\frac{\pi}{4}} \log(1 + \tan x) \, dx \]

2. **Using the substitution \( x = \frac{\pi}{4} - t \):**
\[ I = \int_{0}^{\frac{\pi}{4}} \log(1 + \tan\left(\frac{\pi}{4} - t\right)) \, dt \]
Since \( \tan\left(\frac{\pi}{4} - t\right) = \frac{1 - \tan t}{1 + \tan t} \).

3. **Adding the two integrals:**
\[ 2I = \int_{0}^{\frac{\pi}{4}} \left( \log(1 + \tan x) + \log(1 + \frac{1 - \tan x}{1 + \tan x}) \right) dx \]

4. **Simplifying the integrand:**
\[ 2I = \int_{0}^{\frac{\pi}{4}} \log\left(1 + \tan x + 1 - \tan x\right) \, dx = \int_{0}^{\frac{\pi}{4}} \log(2) \, dx \]

5. **Evaluating the integral:**
\[ 2I = \log(2) \cdot \frac{\pi}{4} \implies I = \frac{\pi}{8} \log(2) \]

### Part (iv): Evaluate \( I = \int_{0}^{\frac{\pi}{2}} \frac{\sin x - \cos x}{1 + \sin x \cos x} \, dx \)

1. **Using the property of definite integrals:**
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\sin x - \cos x}{1 + \sin x \cos x} \, dx \]

2. **Using the substitution \( x = \frac{\pi}{2} - t \):**
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\cos t - \sin t}{1 + \cos t \sin t} \, dt \]

3. **Adding the two integrals:**
\[ 2I = \int_{0}^{\frac{\pi}{2}} \frac{(\sin x - \cos x) + (\cos x - \sin x)}{1 + \sin x \cos x} \, dx = 0 \]

4. **Thus, we have:**
\[ 2I = 0 \implies I = 0 \]

### Summary of Results:
1. \( \int_{0}^{\frac{\pi}{2}} \frac{dx}{1 + \sqrt{\tan x}} = \frac{\pi}{4} \)
2. \( \int_{0}^{\frac{\pi}{2}} \log(\tan x) \, dx = 0 \)
3. \( \int_{0}^{\frac{\pi}{4}} \log(1 + \tan x) \, dx = \frac{\pi}{8} \log(2) \)
4. \( \int_{0}^{\frac{\pi}{2}} \frac{\sin x - \cos x}{1 + \sin x \cos x} \, dx = 0 \)