(i) `int(sec x* cosec x)/(log cot x) dx` (ii) `int tan^(4) x\ dx`

Answer

Let's solve the given problems step by step.

### Part (i): \(\int \frac{\sec x \cdot \csc x}{\log \cot x} \, dx\)

**Step 1: Substitution**
Let \( t = \log \cot x \).

**Hint:** Remember that the logarithm of a quotient can be simplified using properties of logarithms.

**Step 2: Differentiate \( t \)**
Now, differentiate both sides with respect to \( x \):
\[ dt = \frac{d}{dx}(\log \cot x) = \frac{1}{\cot x} \cdot \frac{d}{dx}(\cot x) = \frac{1}{\cot x} \cdot (-\csc^2 x) \, dx \]
This simplifies to:
\[ dt = -\frac{\csc^2 x}{\cot x} \, dx = -\frac{1}{\sin x \cos x} \, dx \]

**Hint:** Use the definitions of sine and cosine to rewrite cotangent and cosecant.

**Step 3: Rewrite in terms of \( t \)**
Rearranging gives:
\[ dx = -\cot x \cdot \sin^2 x \, dt \]
Now, substitute \( \sec x \) and \( \csc x \):
\[ \sec x = \frac{1}{\cos x}, \quad \csc x = \frac{1}{\sin x} \]
Thus, we have:
\[ \sec x \cdot \csc x = \frac{1}{\sin x \cos x} \]

**Step 4: Substitute back into the integral**
Substituting into the integral:
\[ \int \frac{\sec x \cdot \csc x}{\log \cot x} \, dx = \int \frac{1}{\sin x \cos x} \cdot -\cot x \cdot \sin^2 x \, dt \]
This simplifies to:
\[ -\int \frac{\sin x}{\cos x \log \cot x} \, dt = -\int \frac{1}{\log \cot x} \, dt \]

**Step 5: Integrate**
The integral becomes:
\[ -\int \frac{1}{t} \, dt = -\log |t| + C \]
Substituting back \( t = \log \cot x \):
\[ -\log |\log \cot x| + C \]

### Final Answer for Part (i):
\[ -\log |\log \cot x| + C \]

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### Part (ii): \(\int \tan^4 x \, dx\)

**Step 1: Rewrite \(\tan^4 x\)**
We can express \(\tan^4 x\) as:
\[ \tan^4 x = (\tan^2 x)^2 = \sec^2 x - 1 \]

**Hint:** Use the identity \(\tan^2 x = \sec^2 x - 1\).

**Step 2: Substitute**
Thus, we can rewrite the integral:
\[ \int \tan^4 x \, dx = \int (\sec^2 x - 1)^2 \, dx \]

**Step 3: Expand the integrand**
Expanding gives:
\[ \int (\sec^4 x - 2\sec^2 x + 1) \, dx \]

**Step 4: Separate the integrals**
Now we can separate the integrals:
\[ \int \sec^4 x \, dx - 2\int \sec^2 x \, dx + \int 1 \, dx \]

**Step 5: Integrate each part**
1. \(\int \sec^2 x \, dx = \tan x + C\)
2. \(\int 1 \, dx = x + C\)
3. For \(\int \sec^4 x \, dx\), we can use the identity:
\[ \sec^4 x = \tan^2 x + 1 \]
Thus:
\[ \int \sec^4 x \, dx = \tan x + \int \sec^2 x \, dx = \tan x + \tan x = 2\tan x \]

**Step 6: Combine results**
Putting it all together:
\[ \int \tan^4 x \, dx = 2\tan x - 2\tan x + x + C = x + C \]

### Final Answer for Part (ii):
\[ x + C \]

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