To solve the integral $3 \tan^8(x) \sec^4(x) \, dx$, we can apply an appropriate substitution. First, let's rewrite the given expression:

$3 \tan^8(x) \sec^4(x) \, dx = 3 \tan^8(x) \sec^2(x) \sec^2(x) \, dx$

Now, notice that $\sec^2(x) \, dx$ is the derivative of $\tan(x)$. This suggests that we can use the following substitution:

Let: $u = \tan(x)$ Then: $du = \sec^2(x) \, dx$

Now, rewrite the integral in terms of $u$. Using the substitution $u = \tan(x)$, the original expression becomes:

$3 u^8 \sec^2(x) \, du$

Since $du = \sec^2(x) \, dx$, this simplifies further to:

$3 u^8 \, du$

Now, the integral becomes:

$3 \int u^8 \, du$

To solve this, use the power rule of integration:

$\int u^n \, du = \frac{u^{n+1}}{n+1}$

Applying this to $u^8$, we get:

$3 \cdot \frac{u^9}{9} = \frac{u^9}{3}$

Finally, substitute $u = \tan(x)$ back in:

$\frac{\tan^9(x)}{3} + C$

Thus, the integral is:

$\boxed{\frac{\tan^9(x)}{3} + C}$

Would you like more details on any part of this process?

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Here are five related questions:

1. How would you solve integrals involving products of trigonometric functions?
2. Can you explain how to differentiate the result $\frac{\tan^9(x)}{3}$?
3. What is the integral of $\sec^2(x)$?
4. How do you integrate $\tan^n(x)$ for different values of $n$?
5. Why is substitution a useful method for solving integrals?

Tip: When choosing a substitution in integrals, look for patterns that simplify the differential, especially derivatives of trigonometric functions.