Integrate : `int (tan^4 sqrtx sec^2 sqrtx)/sqrtx dx.`

Answer

To solve the integral \( \int \frac{\tan^4(\sqrt{x}) \sec^2(\sqrt{x})}{\sqrt{x}} \, dx \), we will use a substitution method. Here are the steps:

### Step-by-Step Solution:

1. **Substitution**:
Let \( u = \tan(\sqrt{x}) \). Then, we need to find \( du \).
\[ \frac{du}{dx} = \sec^2(\sqrt{x}) \cdot \frac{d}{dx}(\sqrt{x}) = \sec^2(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}} \]
Therefore,
\[ du = \sec^2(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}} \, dx \quad \Rightarrow \quad dx = 2\sqrt{x} \, du \cdot \frac{1}{\sec^2(\sqrt{x})} \]

2. **Rewriting the Integral**:
Substitute \( u \) and \( dx \) into the integral:
\[ \int \frac{\tan^4(\sqrt{x}) \sec^2(\sqrt{x})}{\sqrt{x}} \, dx = \int \frac{u^4 \sec^2(\sqrt{x})}{\sqrt{x}} \cdot 2\sqrt{x} \, du \]
This simplifies to:
\[ 2 \int u^4 \, du \]

3. **Integrating**:
Now, we can integrate \( 2 \int u^4 \, du \):
\[ 2 \cdot \frac{u^5}{5} + C = \frac{2}{5} u^5 + C \]

4. **Back Substitution**:
Substitute back \( u = \tan(\sqrt{x}) \):
\[ \frac{2}{5} \tan^5(\sqrt{x}) + C \]

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