Evaluate the integral.
Integral from nothing to nothing 25 tangent Superscript 5 Baseline x secant Superscript 4 Baseline x dx
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Part 1
Integral from nothing to nothing 25 tangent Superscript 5 Baseline x secant Superscript 4 Baseline x dxequals
  
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We are tasked with evaluating the integral:

\[
\int 25 \tan^5(x) \sec^4(x) \, dx
\]

### Step-by-Step Solution:

1. **Factor out the constant**:
   Since the constant 25 is independent of \( x \), we can factor it out of the integral:

   \[
   25 \int \tan^5(x) \sec^4(x) \, dx
   \]

2. **Rewrite in terms of trigonometric identities**:
   We can rewrite the powers of tangent and secant using trigonometric identities. The most useful identity here is:

   \[
   \sec^2(x) = 1 + \tan^2(x)
   \]

   This will help simplify the integral by relating secant and tangent functions.

3. **Separate one secant-squared term**:
   Recognize that \( \sec^4(x) = \sec^2(x) \cdot \sec^2(x) \). This separation will help simplify the integral using substitution. We rewrite the integral as:

   \[
   25 \int \tan^5(x) \sec^2(x) \cdot \sec^2(x) \, dx
   \]

4. **Substitution**:
   Let \( u = \tan(x) \), so that \( du = \sec^2(x) \, dx \). This transforms the integral into:

   \[
   25 \int u^5 \cdot \sec^2(x) \, dx
   \]

   Since \( du = \sec^2(x) \, dx \), the integral simplifies to:

   \[
   25 \int u^5 \, du
   \]

5. **Integrate**:
   The integral of \( u^5 \) is straightforward:

   \[
   \int u^5 \, du = \frac{u^6}{6}
   \]

   Therefore, the integral becomes:

   \[
   25 \cdot \frac{u^6}{6} = \frac{25}{6} u^6
   \]

6. **Substitute back \( u = \tan(x) \)**:
   Now, substitute back \( u = \tan(x) \) to get the final result:

   \[
   \frac{25}{6} \tan^6(x) + C
   \]

### Final Answer:
\[
\int 25 \tan^5(x) \sec^4(x) \, dx = \frac{25}{6} \tan^6(x) + C
\]

Would you like further clarification on any step or more details?

Here are some related questions to consider:

1. How do we handle integrals involving higher powers of trigonometric functions?
2. What other substitution techniques can simplify integrals like this?
3. Can we apply this process to different powers of tangent and secant functions?
4. How does the secant identity simplify the integration process?
5. What are some typical integral patterns involving tangent and secant?

**Tip:** For integrals involving trigonometric functions raised to powers, try using identities to express them in terms of simpler functions. This often leads to easier substitutions.

Evaluate the integral: Integral from nothing to nothing 25 tangent^5(x) secant^4(x) dx.

Learn how to evaluate the integral of 25 tan^5(x) sec^4(x) with a detailed step-by-step solution. This problem covers key concepts from integral calculus, including the use of trigonometric identities and substitution. Suitable for advanced high school and college-level calculus students.

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