` (i) int sqrt(1+ sin .(x)/(2)dx)`
`(ii) int (1+cos 4x)/(cot x- tan x)dx`

Answer

Let's solve the given integrals step by step.

### Part (i): Evaluate the integral \( \int \sqrt{1 + \frac{\sin x}{2}} \, dx \)

1. **Rewrite the expression under the square root**:
\[ 1 + \frac{\sin x}{2} = \frac{2}{2} + \frac{\sin x}{2} = \frac{2 + \sin x}{2} \]
Thus, we can rewrite the integral as:
\[ \int \sqrt{1 + \frac{\sin x}{2}} \, dx = \int \sqrt{\frac{2 + \sin x}{2}} \, dx = \int \frac{\sqrt{2 + \sin x}}{\sqrt{2}} \, dx = \frac{1}{\sqrt{2}} \int \sqrt{2 + \sin x} \, dx \]

2. **Use the identity for \( \sin x \)**:
We can express \( \sin x \) in terms of \( \sin^2 \) and \( \cos^2 \):
\[ 1 = \sin^2 \frac{x}{4} + \cos^2 \frac{x}{4} \]
Therefore, we can write:
\[ \sqrt{2 + \sin x} = \sqrt{2 + 2\sin\frac{x}{2}\cos\frac{x}{2}} = \sqrt{(\sin\frac{x}{4} + \cos\frac{x}{4})^2} \]

3. **Integrate**:
Now, we can integrate:
\[ \int \sqrt{2 + \sin x} \, dx = \int \left( \sin\frac{x}{4} + \cos\frac{x}{4} \right) \, dx \]
This gives:
\[ = -4\cos\frac{x}{4} + 4\sin\frac{x}{4} + C \]

4. **Final result**:
Therefore, the final result for part (i) is:
\[ \int \sqrt{1 + \frac{\sin x}{2}} \, dx = \frac{1}{\sqrt{2}} \left(-4\cos\frac{x}{4} + 4\sin\frac{x}{4}\right) + C \]

### Part (ii): Evaluate the integral \( \int \frac{1 + \cos 4x}{\cot x - \tan x} \, dx \)

1. **Rewrite the denominator**:
\[ \cot x - \tan x = \frac{\cos x}{\sin x} - \frac{\sin x}{\cos x} = \frac{\cos^2 x - \sin^2 x}{\sin x \cos x} = \frac{\cos 2x}{\sin x \cos x} \]

2. **Rewrite the integral**:
Thus, we can rewrite the integral as:
\[ \int \frac{1 + \cos 4x}{\cot x - \tan x} \, dx = \int \frac{(1 + \cos 4x) \sin x \cos x}{\cos 2x} \, dx \]

3. **Use the identity for \( \cos 4x \)**:
Using the identity \( 1 + \cos 4x = 2\cos^2 2x \):
\[ = \int \frac{2\cos^2 2x \sin x \cos x}{\cos 2x} \, dx = 2 \int \sin x \cos x \cos 2x \, dx \]

4. **Use the double angle identity**:
Using \( \sin 2x = 2\sin x \cos x \):
\[ = \int \sin 2x \cos 2x \, dx \]

5. **Integrate**:
The integral of \( \sin 2x \cos 2x \) can be simplified using the identity:
\[ \sin 2x \cos 2x = \frac{1}{2} \sin 4x \]
Thus:
\[ = \frac{1}{2} \int \sin 4x \, dx = -\frac{1}{8} \cos 4x + C \]

6. **Final result**:
Therefore, the final result for part (ii) is:
\[ \int \frac{1 + \cos 4x}{\cot x - \tan x} \, dx = -\frac{1}{8} \cos 4x + C \]