(i) `int(1)/((1+cos2x))dx` , (ii) `int (1)/((1-cos2x)) dx`

Answer

Let's solve the integrals step by step.

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

1. **Use the identity for \(\cos 2x\)**:
\[ \cos 2x = 2\cos^2 x - 1 \]
Therefore,
\[ 1 + \cos 2x = 1 + (2\cos^2 x - 1) = 2\cos^2 x \]

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

3. **Integrate \(\sec^2 x\)**:
The integral of \(\sec^2 x\) is:
\[ \int \sec^2 x \, dx = \tan x + C \]

4. **Combine the results**:
Thus,
\[ \frac{1}{2} \int \sec^2 x \, dx = \frac{1}{2} (\tan x + C) = \frac{1}{2} \tan x + C \]

### Final Answer for Part (i):
\[ \int \frac{1}{1 + \cos 2x} \, dx = \frac{1}{2} \tan x + C \]

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

1. **Use the identity for \(\cos 2x\)**:
\[ \cos 2x = 1 - 2\sin^2 x \]
Therefore,
\[ 1 - \cos 2x = 1 - (1 - 2\sin^2 x) = 2\sin^2 x \]

2. **Rewrite the integral**:
\[ \int \frac{1}{1 - \cos 2x} \, dx = \int \frac{1}{2\sin^2 x} \, dx = \frac{1}{2} \int \csc^2 x \, dx \]

3. **Integrate \(\csc^2 x\)**:
The integral of \(\csc^2 x\) is:
\[ \int \csc^2 x \, dx = -\cot x + C \]

4. **Combine the results**:
Thus,
\[ \frac{1}{2} \int \csc^2 x \, dx = \frac{1}{2} (-\cot x + C) = -\frac{1}{2} \cot x + C \]

### Final Answer for Part (ii):
\[ \int \frac{1}{1 - \cos 2x} \, dx = -\frac{1}{2} \cot x + C \]

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