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

Explanation

To solve the integral \( \int \tan^2 x \sec^4 x \, dx \), we will follow a systematic approach.

### Step-by-step Solution:

1. **Set up the integral:**
\[ I = \int \tan^2 x \sec^4 x \, dx \]

2. **Use substitution:**
Let \( t = \tan x \). Then, the derivative of \( t \) with respect to \( x \) is:
\[ \frac{dt}{dx} = \sec^2 x \quad \Rightarrow \quad dt = \sec^2 x \, dx \]

3. **Rewrite the integral in terms of \( t \):**
Since \( \tan^2 x = t^2 \) and \( \sec^4 x = \sec^2 x \cdot \sec^2 x \), we can express the integral as:
\[ I = \int t^2 \sec^4 x \, dx = \int t^2 \sec^2 x \cdot \sec^2 x \, dx \]
Now substituting \( dx \) with \( dt \):
\[ I = \int t^2 \sec^2 x \, dt \]

4. **Relate \( \sec^2 x \) to \( t \):**
From the identity \( \tan^2 x + 1 = \sec^2 x \), we have:
\[ \sec^2 x = t^2 + 1 \]
Thus, we can rewrite the integral:
\[ I = \int t^2 (t^2 + 1) \, dt \]

5. **Expand the integrand:**
\[ I = \int (t^4 + t^2) \, dt \]

6. **Integrate term by term:**
\[ I = \int t^4 \, dt + \int t^2 \, dt = \frac{t^5}{5} + \frac{t^3}{3} + C \]

7. **Substitute back for \( t \):**
Recall that \( t = \tan x \):
\[ I = \frac{\tan^5 x}{5} + \frac{\tan^3 x}{3} + C \]

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