Evaluate: (i) `int1/(sqrt(1+cos2x))\ dx` (ii) `int1/(sqrt(1-cosx))\ dx`

Answer

To solve the integrals given in the question, we will evaluate each one step by step.

### (i) Evaluate \( \int \frac{1}{\sqrt{1 + \cos 2x}} \, dx \)

1. **Use the identity for \( \cos 2x \)**:
\[ 1 + \cos 2x = 2 \cos^2 x \]
Therefore, we can rewrite the integral as:
\[ \int \frac{1}{\sqrt{2 \cos^2 x}} \, dx \]

2. **Simplify the integral**:
\[ \int \frac{1}{\sqrt{2} \cos x} \, dx = \frac{1}{\sqrt{2}} \int \sec x \, dx \]

3. **Integrate \( \sec x \)**:
The integral of \( \sec x \) is:
\[ \int \sec x \, dx = \ln |\sec x + \tan x| + C \]
Thus, we have:
\[ \frac{1}{\sqrt{2}} \left( \ln |\sec x + \tan x| + C \right) \]

4. **Final result for the first integral**:
\[ \int \frac{1}{\sqrt{1 + \cos 2x}} \, dx = \frac{1}{\sqrt{2}} \ln |\sec x + \tan x| + C \]

### (ii) Evaluate \( \int \frac{1}{\sqrt{1 - \cos 2x}} \, dx \)

1. **Use the identity for \( \cos 2x \)**:
\[ 1 - \cos 2x = 2 \sin^2 x \]
Therefore, we can rewrite the integral as:
\[ \int \frac{1}{\sqrt{2 \sin^2 x}} \, dx \]

2. **Simplify the integral**:
\[ \int \frac{1}{\sqrt{2} \sin x} \, dx = \frac{1}{\sqrt{2}} \int \csc x \, dx \]

3. **Integrate \( \csc x \)**:
The integral of \( \csc x \) is:
\[ \int \csc x \, dx = \ln |\csc x - \cot x| + C \]
Thus, we have:
\[ \frac{1}{\sqrt{2}} \left( \ln |\csc x - \cot x| + C \right) \]

4. **Final result for the second integral**:
\[ \int \frac{1}{\sqrt{1 - \cos 2x}} \, dx = \frac{1}{\sqrt{2}} \ln |\csc x - \cot x| + C \]

### Summary of Results
- (i) \( \int \frac{1}{\sqrt{1 + \cos 2x}} \, dx = \frac{1}{\sqrt{2}} \ln |\sec x + \tan x| + C \)
- (ii) \( \int \frac{1}{\sqrt{1 - \cos 2x}} \, dx = \frac{1}{\sqrt{2}} \ln |\csc x - \cot x| + C \)