`int tan^(2)xsec^(4)x dx`

Explanation

To solve the integral \( \int \tan^2 x \sec^4 x \, dx \), we can follow these steps:

### Step 1: Rewrite the integral
We know that \( \sec^2 x = 1 + \tan^2 x \). Therefore, we can express \( \sec^4 x \) as:
\[ \sec^4 x = \sec^2 x \cdot \sec^2 x = (1 + \tan^2 x) \sec^2 x \]
Thus, we can rewrite the integral as:
\[ \int \tan^2 x \sec^4 x \, dx = \int \tan^2 x (1 + \tan^2 x) \sec^2 x \, dx \]

### Step 2: Substitute \( u = \tan x \)
Let \( u = \tan x \). Then, the derivative \( du = \sec^2 x \, dx \). This means \( dx = \frac{du}{\sec^2 x} \).

### Step 3: Change the variable in the integral
Substituting \( u \) into the integral, we have:
\[ \int u^2 (1 + u^2) \sec^2 x \, dx = \int u^2 (1 + u^2) \, du \]
Now, we can expand the integrand:
\[ \int u^2 + u^4 \, du \]

### Step 4: Integrate term by term
Now we integrate each term separately:
\[ \int u^2 \, du + \int u^4 \, du = \frac{u^3}{3} + \frac{u^5}{5} + C \]

### Step 5: Substitute back \( u = \tan x \)
Now we substitute back \( u = \tan x \):
\[ \frac{\tan^3 x}{3} + \frac{\tan^5 x}{5} + C \]

### Final Answer
Thus, the final answer is:
\[ \int \tan^2 x \sec^4 x \, dx = \frac{\tan^3 x}{3} + \frac{\tan^5 x}{5} + C \]

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