(i) `int_0^(pi/2) sin^2 x dx`
(ii) `int_0^(pi//2) cos^2 x dx`

Answer

To solve the integrals \( \int_0^{\frac{\pi}{2}} \sin^2 x \, dx \) and \( \int_0^{\frac{\pi}{2}} \cos^2 x \, dx \), we will use trigonometric identities to simplify the integrands.

### Step-by-Step Solution

**(i) Calculate \( \int_0^{\frac{\pi}{2}} \sin^2 x \, dx \)**

1. **Use the identity for \( \sin^2 x \)**:
\[ \sin^2 x = \frac{1 - \cos 2x}{2} \]
Therefore, we can rewrite the integral as:
\[ \int_0^{\frac{\pi}{2}} \sin^2 x \, dx = \int_0^{\frac{\pi}{2}} \frac{1 - \cos 2x}{2} \, dx \]

2. **Split the integral**:
\[ = \frac{1}{2} \int_0^{\frac{\pi}{2}} (1 - \cos 2x) \, dx = \frac{1}{2} \left( \int_0^{\frac{\pi}{2}} 1 \, dx - \int_0^{\frac{\pi}{2}} \cos 2x \, dx \right) \]

3. **Calculate the first integral**:
\[ \int_0^{\frac{\pi}{2}} 1 \, dx = \left[ x \right]_0^{\frac{\pi}{2}} = \frac{\pi}{2} \]

4. **Calculate the second integral**:
\[ \int_0^{\frac{\pi}{2}} \cos 2x \, dx = \left[ \frac{\sin 2x}{2} \right]_0^{\frac{\pi}{2}} = \frac{\sin \pi}{2} - \frac{\sin 0}{2} = 0 - 0 = 0 \]

5. **Combine the results**:
\[ \int_0^{\frac{\pi}{2}} \sin^2 x \, dx = \frac{1}{2} \left( \frac{\pi}{2} - 0 \right) = \frac{\pi}{4} \]

**(ii) Calculate \( \int_0^{\frac{\pi}{2}} \cos^2 x \, dx \)**

1. **Use the identity for \( \cos^2 x \)**:
\[ \cos^2 x = \frac{1 + \cos 2x}{2} \]
Therefore, we can rewrite the integral as:
\[ \int_0^{\frac{\pi}{2}} \cos^2 x \, dx = \int_0^{\frac{\pi}{2}} \frac{1 + \cos 2x}{2} \, dx \]

2. **Split the integral**:
\[ = \frac{1}{2} \int_0^{\frac{\pi}{2}} (1 + \cos 2x) \, dx = \frac{1}{2} \left( \int_0^{\frac{\pi}{2}} 1 \, dx + \int_0^{\frac{\pi}{2}} \cos 2x \, dx \right) \]

3. **Calculate the first integral** (already done):
\[ \int_0^{\frac{\pi}{2}} 1 \, dx = \frac{\pi}{2} \]

4. **Calculate the second integral** (already done):
\[ \int_0^{\frac{\pi}{2}} \cos 2x \, dx = 0 \]

5. **Combine the results**:
\[ \int_0^{\frac{\pi}{2}} \cos^2 x \, dx = \frac{1}{2} \left( \frac{\pi}{2} + 0 \right) = \frac{\pi}{4} \]

### Final Answers
- \( \int_0^{\frac{\pi}{2}} \sin^2 x \, dx = \frac{\pi}{4} \)
- \( \int_0^{\frac{\pi}{2}} \cos^2 x \, dx = \frac{\pi}{4} \)