(i) `int_0^(pi//4) sec x dx`
(ii) `int_(pi//6)^(pi//4) cosec x dx`

Answer

Let's solve the given integrals step by step.

### (i) Evaluate \( \int_0^{\frac{\pi}{4}} \sec x \, dx \)

1. **Substitution**: We know that \( \sec x = \frac{1}{\cos x} \). To integrate \( \sec x \), we can use the identity:
\[ \sec x = \frac{1}{\cos x} = \frac{\sec x + \tan x}{\sec x + \tan x} \cdot \sec x \]
This allows us to rewrite the integral:
\[ \int \sec x \, dx = \int \frac{\sec x (\sec x + \tan x)}{\sec x + \tan x} \, dx \]

2. **Differentiation**: Let \( t = \sec x + \tan x \). Then, differentiating both sides gives:
\[ dt = (\sec x \tan x + \sec^2 x) \, dx \]
Thus, \( dx = \frac{dt}{\sec x \tan x + \sec^2 x} \).

3. **Change of Limits**: When \( x = 0 \):
\[ t = \sec(0) + \tan(0) = 1 + 0 = 1 \]
When \( x = \frac{\pi}{4} \):
\[ t = \sec\left(\frac{\pi}{4}\right) + \tan\left(\frac{\pi}{4}\right) = \sqrt{2} + 1 \]

4. **Substituting Back**: The integral now becomes:
\[ \int_1^{1+\sqrt{2}} \frac{1}{t} \, dt \]

5. **Integrating**: The integral of \( \frac{1}{t} \) is:
\[ \ln |t| \Big|_1^{1+\sqrt{2}} = \ln(1+\sqrt{2}) - \ln(1) = \ln(1+\sqrt{2}) \]

Thus, the result for the first integral is:
\[ \int_0^{\frac{\pi}{4}} \sec x \, dx = \ln(1+\sqrt{2}) \]

### (ii) Evaluate \( \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \csc x \, dx \)

1. **Substitution**: We know that \( \csc x = \frac{1}{\sin x} \). To integrate \( \csc x \), we can use the identity:
\[ \csc x = \frac{\csc x (\csc x + \cot x)}{\csc x + \cot x} \cdot \csc x \]

2. **Differentiation**: Let \( t = \csc x + \cot x \). Then, differentiating gives:
\[ dt = (-\csc x \cot x - \csc^2 x) \, dx \]
Thus, \( dx = \frac{dt}{-\csc x \cot x - \csc^2 x} \).

3. **Change of Limits**: When \( x = \frac{\pi}{6} \):
\[ t = \csc\left(\frac{\pi}{6}\right) + \cot\left(\frac{\pi}{6}\right) = 2 + \sqrt{3} \]
When \( x = \frac{\pi}{4} \):
\[ t = \csc\left(\frac{\pi}{4}\right) + \cot\left(\frac{\pi}{4}\right) = \sqrt{2} + 1 \]

4. **Substituting Back**: The integral now becomes:
\[ -\int_{2+\sqrt{3}}^{\sqrt{2}+1} \frac{1}{t} \, dt \]

5. **Integrating**: The integral of \( \frac{1}{t} \) is:
\[ -\left( \ln |t| \Big|_{2+\sqrt{3}}^{\sqrt{2}+1} \right) = -\left( \ln(\sqrt{2}+1) - \ln(2+\sqrt{3}) \right) \]
This simplifies to:
\[ \ln(2+\sqrt{3}) - \ln(\sqrt{2}+1) \]
Using the property of logarithms:
\[ = \ln\left(\frac{2+\sqrt{3}}{\sqrt{2}+1}\right) \]

Thus, the result for the second integral is:
\[ \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \csc x \, dx = \ln\left(\frac{2+\sqrt{3}}{\sqrt{2}+1}\right) \]

### Summary of Results:
1. \( \int_0^{\frac{\pi}{4}} \sec x \, dx = \ln(1+\sqrt{2}) \)
2. \( \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \csc x \, dx = \ln\left(\frac{2+\sqrt{3}}{\sqrt{2}+1}\right) \)