Evaluate:
`int sqrt(1+cos2x) dx`
(ii) `int sqrt(1+sin2x) dx.`

Answer

Let's evaluate the integrals step by step.

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

1. **Use the identity for cosine**: We know that \( \cos 2x = 2\cos^2 x - 1 \). Therefore, we can rewrite \( 1 + \cos 2x \) as:
\[ 1 + \cos 2x = 1 + (2\cos^2 x - 1) = 2\cos^2 x \]

2. **Take the square root**: Now, we can take the square root:
\[ \sqrt{1 + \cos 2x} = \sqrt{2\cos^2 x} = \sqrt{2} \cdot \cos x \]

3. **Set up the integral**: Substitute this back into the integral:
\[ \int \sqrt{1 + \cos 2x} \, dx = \int \sqrt{2} \cdot \cos x \, dx \]

4. **Integrate**: The integral of \( \cos x \) is \( \sin x \):
\[ \int \sqrt{2} \cdot \cos x \, dx = \sqrt{2} \cdot \sin x + C \]

Thus, the solution for part (i) is:
\[ \int \sqrt{1 + \cos 2x} \, dx = \sqrt{2} \sin x + C \]

### Part (ii): Evaluate \( \int \sqrt{1 + \sin 2x} \, dx \)

1. **Use the identity for sine**: We know that \( \sin 2x = 2\sin x \cos x \). Therefore, we can rewrite \( 1 + \sin 2x \) as:
\[ 1 + \sin 2x = 1 + 2\sin x \cos x \]

2. **Recognize a perfect square**: We can express this as:
\[ 1 + \sin 2x = (\sin x + \cos x)^2 \]
This is because:
\[ (\sin x + \cos x)^2 = \sin^2 x + 2\sin x \cos x + \cos^2 x = 1 + 2\sin x \cos x \]

3. **Take the square root**: Now, we can take the square root:
\[ \sqrt{1 + \sin 2x} = \sin x + \cos x \]

4. **Set up the integral**: Substitute this back into the integral:
\[ \int \sqrt{1 + \sin 2x} \, dx = \int (\sin x + \cos x) \, dx \]

5. **Integrate**: The integral of \( \sin x \) is \( -\cos x \) and the integral of \( \cos x \) is \( \sin x \):
\[ \int (\sin x + \cos x) \, dx = -\cos x + \sin x + C \]

Thus, the solution for part (ii) is:
\[ \int \sqrt{1 + \sin 2x} \, dx = -\cos x + \sin x + C \]

### Summary of Solutions
1. \( \int \sqrt{1 + \cos 2x} \, dx = \sqrt{2} \sin x + C \)
2. \( \int \sqrt{1 + \sin 2x} \, dx = -\cos x + \sin x + C \)