To find the antiderivative of $\sec^3(\pi x)$, we can use a standard reduction formula for integrals involving $\sec^3(x)$. The general approach for this kind of integral is:

$\int \sec^3(x) \, dx = \frac{1}{2} \sec(x) \tan(x) + \frac{1}{2} \int \sec(x) \, dx$

In your case, the argument is $\pi x$ rather than $x$, so we need to handle that.

The integral you're asking about is:

$\int \sec^3(\pi x) \, dx$

Step-by-step solution:

1. Substitute $u = \pi x$, so $du = \pi \, dx$, or equivalently $dx = \frac{du}{\pi}$.

   The integral becomes:

   $\int \sec^3(\pi x) \, dx = \frac{1}{\pi} \int \sec^3(u) \, du$

2. Now, apply the reduction formula for $\sec^3(u)$:

   $\int \sec^3(u) \, du = \frac{1}{2} \sec(u) \tan(u) + \frac{1}{2} \int \sec(u) \, du$

   So, we have:

   $\frac{1}{\pi} \left( \frac{1}{2} \sec(u) \tan(u) + \frac{1}{2} \int \sec(u) \, du \right)$

3. The integral of $\sec(u)$ is straightforward:

   $\int \sec(u) \, du = \ln | \sec(u) + \tan(u) | + C$

4. Substituting $u = \pi x$ back into the expression:

   $\frac{1}{\pi} \left( \frac{1}{2} \sec(\pi x) \tan(\pi x) + \frac{1}{2} \ln | \sec(\pi x) + \tan(\pi x) | + C \right)$

Final Answer:

$\int \sec^3(\pi x) \, dx = \frac{1}{2\pi} \sec(\pi x) \tan(\pi x) + \frac{1}{2\pi} \ln | \sec(\pi x) + \tan(\pi x) | + C$

Let me know if you'd like more details or further clarification!

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Here are some related questions you might find interesting:

1. What is the integral of $\sec^2(x)$?
2. How can I solve integrals with $\tan(x)$ or $\cos(x)$ as the primary function?
3. How can the reduction formula for $\sec^3(x)$ be derived?
4. What is the general strategy for handling trigonometric integrals involving higher powers of secant or tangent?
5. How do I evaluate definite integrals with trigonometric functions?

Tip: For integrals involving secant and tangent, often using trigonometric identities and reduction formulas can simplify the process considerably.