`(i) int(cos^(3) x+ sin^(3) x)/(sin^(2) x.cos ^(2) x)dx " "(ii) int(cos2x)/(cos^(2) x sin^(2)x) dx`

Answer

Let's solve the given integrals step by step.

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

\[ \int \frac{\cos^3 x + \sin^3 x}{\sin^2 x \cos^2 x} \, dx \]

**Step 1: Simplify the integrand**

Using the identity for the sum of cubes, we can rewrite \(\cos^3 x + \sin^3 x\) as:

\[ \cos^3 x + \sin^3 x = (\cos x + \sin x)(\cos^2 x - \cos x \sin x + \sin^2 x) \]

Since \(\cos^2 x + \sin^2 x = 1\), we have:

\[ \cos^3 x + \sin^3 x = (\cos x + \sin x)(1 - \cos x \sin x) \]

Thus, the integral becomes:

\[ \int \frac{(\cos x + \sin x)(1 - \cos x \sin x)}{\sin^2 x \cos^2 x} \, dx \]

**Step 2: Split the integral**

We can split the integral into two parts:

\[ \int \frac{\cos x + \sin x}{\sin^2 x \cos^2 x} \, dx - \int \frac{(\cos x + \sin x) \cos x \sin x}{\sin^2 x \cos^2 x} \, dx \]

This simplifies to:

\[ \int \frac{\cos x}{\sin^2 x \cos^2 x} \, dx + \int \frac{\sin x}{\sin^2 x \cos^2 x} \, dx \]

**Step 3: Rewrite the integrals**

The first integral can be rewritten as:

\[ \int \frac{1}{\sin^2 x} \cdot \frac{1}{\cos x} \, dx = \int \cot x \csc x \, dx \]

The second integral can be rewritten as:

\[ \int \frac{1}{\cos^2 x} \cdot \frac{1}{\sin x} \, dx = \int \tan x \sec x \, dx \]

**Step 4: Evaluate the integrals**

The integral \(\int \cot x \csc x \, dx\) evaluates to \(-\csc x\), and the integral \(\int \tan x \sec x \, dx\) evaluates to \(\sec x\).

Thus, we have:

\[ -\csc x + \sec x + C \]

### Final Answer for Part (i):

\[ \int \frac{\cos^3 x + \sin^3 x}{\sin^2 x \cos^2 x} \, dx = \sec x - \csc x + C \]

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

\[ \int \frac{\cos 2x}{\cos^2 x \sin^2 x} \, dx \]

**Step 1: Use the double angle identity**

We know that:

\[ \cos 2x = 2 \cos^2 x - 1 \]

Thus, we can rewrite the integral as:

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

**Step 2: Split the integral**

This can be split into two separate integrals:

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

This simplifies to:

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

**Step 3: Rewrite the integrals**

The first integral can be rewritten as:

\[ 2 \int \csc^2 x \, dx \]

The second integral can be rewritten as:

\[ \int \sec^2 x \, dx \]

**Step 4: Evaluate the integrals**

The integral \(\int \csc^2 x \, dx\) evaluates to \(-\cot x\), and the integral \(\int \sec^2 x \, dx\) evaluates to \(\tan x\).

Thus, we have:

\[ 2(-\cot x) - \tan x + C = -2 \cot x - \tan x + C \]

### Final Answer for Part (ii):

\[ \int \frac{\cos 2x}{\cos^2 x \sin^2 x} \, dx = -2 \cot x - \tan x + C \]

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