To solve the integral \(\int 9 \tan(x) \sec^3(x) \, dx\), we can use the substitution method and recognize patterns in the integral.

Let's start by rewriting the integral:

\[
\int 9 \tan(x) \sec^3(x) \, dx
\]

We can use the substitution \( u = \sec(x) \), which means \( du = \sec(x) \tan(x) \, dx \).

Given this substitution, we can rewrite the integral as:

\[
9 \int \sec(x) \cdot \sec(x) \tan(x) \, dx
\]

Since \( \sec(x) \tan(x) \, dx = du \), the integral becomes:

\[
9 \int u^2 \, du
\]

The integral of \( u^2 \) with respect to \( u \) is:

\[
\frac{u^3}{3} + C
\]

Substituting back \( u = \sec(x) \), we get:

\[
\frac{9}{3} \sec^3(x) + C = 3 \sec^3(x) + C
\]

Thus, the solution to the integral is:

\[
\int 9 \tan(x) \sec^3(x) \, dx = 3 \sec^3(x) + C
\]

Would you like more details or have any questions?

Here are 5 related questions for further exploration:
1. How would you approach the integral \(\int \tan^2(x) \sec(x) \, dx\)?
2. What is the integral of \(\int \sec^4(x) \, dx\)?
3. How does the substitution method differ for different trigonometric integrals?
4. What are the integral properties of \(\sec(x)\) and \(\tan(x)\)?
5. How do you integrate functions involving both \(\tan(x)\) and \(\sec(x)\) raised to different powers?

**Tip:** When integrating functions involving \(\sec(x)\) and \(\tan(x)\), look for ways to express parts of the integrand in terms of derivatives of known functions, like \(\sec(x)\) and \(\tan(x)\), to simplify the process.

Learn how to solve the integral ∫ 9 tan(x) sec^3(x) dx step-by-step with a detailed solution using the substitution method. This problem covers advanced topics in integral calculus and trigonometric integrals.

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