Evaluate the following integrals.
`int tanx sec^(6) x dx `

Answer

To evaluate the integral \( \int \tan x \sec^6 x \, dx \), we can follow these steps:

### Step 1: Rewrite the integral
We start with the integral:
\[ \int \tan x \sec^6 x \, dx \]

### Step 2: Use substitution
We know that the derivative of \( \sec x \) is \( \sec x \tan x \). This suggests that we can use \( u = \sec x \) as our substitution. Thus, we have:
\[ du = \sec x \tan x \, dx \]
This means:
\[ dx = \frac{du}{\sec x \tan x} \]

### Step 3: Rewrite \( \tan x \) and \( \sec^6 x \) in terms of \( u \)
From our substitution \( u = \sec x \), we can express \( \tan x \) as:
\[ \tan x = \sqrt{\sec^2 x - 1} = \sqrt{u^2 - 1} \]
Thus, we can rewrite the integral as:
\[ \int \tan x \sec^6 x \, dx = \int \sqrt{u^2 - 1} \cdot u^6 \cdot \frac{du}{\sec x \tan x} \]

### Step 4: Simplify the integral
Since \( \sec x = u \) and \( \tan x = \frac{\sqrt{u^2 - 1}}{u} \), we can substitute these back into the integral:
\[ \int \tan x \sec^6 x \, dx = \int \sqrt{u^2 - 1} \cdot u^6 \cdot \frac{du}{u \cdot \frac{\sqrt{u^2 - 1}}{u}} = \int u^5 \, du \]

### Step 5: Integrate
Now we can integrate:
\[ \int u^5 \, du = \frac{u^6}{6} + C \]

### Step 6: Substitute back
Now we substitute back \( u = \sec x \):
\[ \frac{\sec^6 x}{6} + C \]

### Final Answer
Thus, the integral evaluates to:
\[ \int \tan x \sec^6 x \, dx = \frac{\sec^6 x}{6} + C \]

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