(i) `int 1/ (cos^4x+sin^4x)dx`
(ii) `int 1/(sin^4x+sin^2x cos^2x+cos^4x) dx`

Answer

Let's solve the given integrals step by step.

### Part (i):
We need to evaluate the integral

\[ I = \int \frac{1}{\cos^4 x + \sin^4 x} \, dx. \]

**Step 1: Simplifying the denominator**

We can rewrite \(\cos^4 x + \sin^4 x\) using the identity:

\[ \cos^4 x + \sin^4 x = (\cos^2 x + \sin^2 x)^2 - 2\cos^2 x \sin^2 x = 1 - 2\cos^2 x \sin^2 x. \]

Thus,

\[ I = \int \frac{1}{1 - 2\cos^2 x \sin^2 x} \, dx. \]

**Step 2: Using the double angle identity**

Using the identity \(\sin^2 x = \frac{1 - \cos(2x)}{2}\) and \(\cos^2 x = \frac{1 + \cos(2x)}{2}\), we can express \(2\cos^2 x \sin^2 x\) as:

\[ 2\cos^2 x \sin^2 x = 2 \cdot \frac{1 + \cos(2x)}{2} \cdot \frac{1 - \cos(2x)}{2} = \frac{1 - \cos^2(2x)}{2} = \frac{\sin^2(2x)}{2}. \]

Thus,

\[ I = \int \frac{1}{1 - \frac{\sin^2(2x)}{2}} \, dx. \]

**Step 3: Further simplification**

This can be rewritten as:

\[ I = \int \frac{2}{2 - \sin^2(2x)} \, dx. \]

**Step 4: Substitution**

Let \(u = 2x\), then \(du = 2dx\) or \(dx = \frac{du}{2}\). The integral becomes:

\[ I = \int \frac{2}{2 - \sin^2 u} \cdot \frac{du}{2} = \int \frac{1}{2 - \sin^2 u} \, du. \]

**Step 5: Using the identity for integration**

Now we can use the formula for integrating \(\frac{1}{a - b \sin^2 u}\):

\[ \int \frac{1}{a - b \sin^2 u} \, du = \frac{1}{\sqrt{a(a+b)}} \tan^{-1} \left( \sqrt{\frac{b}{a}} \tan u \right) + C. \]

In our case, \(a = 2\) and \(b = 1\):

\[ I = \frac{1}{\sqrt{2(2+1)}} \tan^{-1} \left( \sqrt{\frac{1}{2}} \tan u \right) + C = \frac{1}{\sqrt{6}} \tan^{-1} \left( \frac{1}{\sqrt{2}} \tan(2x) \right) + C. \]

### Part (ii):
Now we evaluate the integral

\[ J = \int \frac{1}{\sin^4 x + \sin^2 x \cos^2 x + \cos^4 x} \, dx. \]

**Step 1: Simplifying the denominator**

We can rewrite \(\sin^4 x + \sin^2 x \cos^2 x + \cos^4 x\) as:

\[ \sin^4 x + \cos^4 x + \sin^2 x \cos^2 x = (\sin^2 x + \cos^2 x)^2 - \sin^2 x \cos^2 x = 1 - \sin^2 x \cos^2 x. \]

Thus,

\[ J = \int \frac{1}{1 - \sin^2 x \cos^2 x} \, dx. \]

**Step 2: Using the double angle identity**

Using the identity \(\sin^2 x \cos^2 x = \frac{1}{4} \sin^2(2x)\):

\[ J = \int \frac{1}{1 - \frac{1}{4} \sin^2(2x)} \, dx. \]

**Step 3: Further simplification**

This can be rewritten as:

\[ J = \int \frac{4}{4 - \sin^2(2x)} \, dx. \]

**Step 4: Substitution**

Let \(u = 2x\), then \(du = 2dx\) or \(dx = \frac{du}{2}\). The integral becomes:

\[ J = \int \frac{4}{4 - \sin^2 u} \cdot \frac{du}{2} = 2 \int \frac{1}{4 - \sin^2 u} \, du. \]

**Step 5: Using the identity for integration**

Using the formula for integrating \(\frac{1}{a - b \sin^2 u}\):

In our case, \(a = 4\) and \(b = 1\):

\[ J = 2 \cdot \frac{1}{\sqrt{4(4+1)}} \tan^{-1} \left( \sqrt{\frac{1}{4}} \tan u \right) + C = \frac{2}{\sqrt{20}} \tan^{-1} \left( \frac{1}{2} \tan(2x) \right) + C. \]

### Final Answers:
1. For the first integral:
\[ I = \frac{1}{\sqrt{6}} \tan^{-1} \left( \frac{1}{\sqrt{2}} \tan(2x) \right) + C. \]

2. For the second integral:
\[ J = \frac{2}{\sqrt{20}} \tan^{-1} \left( \frac{1}{2} \tan(2x) \right) + C. \]