Evaluate (i) `int_0^(pi//4) sin x cos x dx` (ii) `int_0^(pi//2) (1 + cos x)^(1//2) dx` (iii) `int_0^(pi//2) (1 + sin x)^(1//2) dx` (iv) `int_0^(pi//4) (1 -cos 2x)^(1//2)dx`

Explanation

Let's solve the given integrals step by step.

### Part (i): Evaluate \( \int_0^{\frac{\pi}{4}} \sin x \cos x \, dx \)

1. **Substitution**: Let \( u = \sin x \). Then, \( du = \cos x \, dx \).
2. **Change of limits**: When \( x = 0 \), \( u = \sin(0) = 0 \). When \( x = \frac{\pi}{4} \), \( u = \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \).
3. **Rewrite the integral**:
\[ \int_0^{\frac{\pi}{4}} \sin x \cos x \, dx = \int_0^{\frac{1}{\sqrt{2}}} u \, du \]
4. **Integrate**:
\[ \int u \, du = \frac{u^2}{2} \Big|_0^{\frac{1}{\sqrt{2}}} = \frac{\left(\frac{1}{\sqrt{2}}\right)^2}{2} - \frac{0^2}{2} = \frac{\frac{1}{2}}{2} = \frac{1}{4} \]

### Part (ii): Evaluate \( \int_0^{\frac{\pi}{2}} (1 + \cos x)^{\frac{1}{2}} \, dx \)

1. **Rewrite the integral**:
\[ \int_0^{\frac{\pi}{2}} (1 + \cos x)^{\frac{1}{2}} \, dx \]
2. **Use the identity**: \( \cos x = 2\cos^2\left(\frac{x}{2}\right) - 1 \) leads to \( 1 + \cos x = 2\cos^2\left(\frac{x}{2}\right) \).
3. **Substitute**:
\[ (1 + \cos x)^{\frac{1}{2}} = \sqrt{2} \cos\left(\frac{x}{2}\right) \]
4. **Integral becomes**:
\[ \sqrt{2} \int_0^{\frac{\pi}{2}} \cos\left(\frac{x}{2}\right) \, dx \]
5. **Change of variable**: Let \( t = \frac{x}{2} \), then \( dx = 2 \, dt \) and the limits change from \( 0 \) to \( \frac{\pi}{4} \):
\[ = 2\sqrt{2} \int_0^{\frac{\pi}{4}} \cos t \, dt \]
6. **Integrate**:
\[ = 2\sqrt{2} \left[\sin t\right]_0^{\frac{\pi}{4}} = 2\sqrt{2} \left(\frac{1}{\sqrt{2}} - 0\right) = 2 \]

### Part (iii): Evaluate \( \int_0^{\frac{\pi}{2}} (1 + \sin x)^{\frac{1}{2}} \, dx \)

1. **Rewrite the integral**:
\[ \int_0^{\frac{\pi}{2}} (1 + \sin x)^{\frac{1}{2}} \, dx \]
2. **Use the identity**: \( 1 + \sin x = \left(\sin\left(\frac{x}{2}\right) + \cos\left(\frac{x}{2}\right)\right)^2 \).
3. **Integral becomes**:
\[ \int_0^{\frac{\pi}{2}} \left(\sin\left(\frac{x}{2}\right) + \cos\left(\frac{x}{2}\right)\right) \, dx \]
4. **Change of variable**: Let \( t = \frac{x}{2} \), then \( dx = 2 \, dt \) and the limits change from \( 0 \) to \( \frac{\pi}{4} \):
\[ = 2 \int_0^{\frac{\pi}{4}} \left(\sin t + \cos t\right) \, dt \]
5. **Integrate**:
\[ = 2 \left[-\cos t + \sin t\right]_0^{\frac{\pi}{4}} = 2 \left[-\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} - (-1)\right] = 2 \left(1 - 0\right) = 2 \]

### Part (iv): Evaluate \( \int_0^{\frac{\pi}{4}} (1 - \cos 2x)^{\frac{1}{2}} \, dx \)

1. **Use the identity**: \( 1 - \cos 2x = 2\sin^2 x \).
2. **Rewrite the integral**:
\[ \int_0^{\frac{\pi}{4}} (1 - \cos 2x)^{\frac{1}{2}} \, dx = \int_0^{\frac{\pi}{4}} \sqrt{2} \sin x \, dx \]
3. **Integrate**:
\[ = \sqrt{2} \left[-\cos x\right]_0^{\frac{\pi}{4}} = \sqrt{2} \left[-\frac{1}{\sqrt{2}} + 1\right] = \sqrt{2} \left(1 - \frac{1}{\sqrt{2}}\right) = \sqrt{2} - 1 \]

### Summary of Results
1. \( \int_0^{\frac{\pi}{4}} \sin x \cos x \, dx = \frac{1}{4} \)
2. \( \int_0^{\frac{\pi}{2}} (1 + \cos x)^{\frac{1}{2}} \, dx = 2 \)
3. \( \int_0^{\frac{\pi}{2}} (1 + \sin x)^{\frac{1}{2}} \, dx = 2 \)
4. \( \int_0^{\frac{\pi}{4}} (1 - \cos 2x)^{\frac{1}{2}} \, dx = \sqrt{2} - 1 \)