The Maclaurin series for the antiderivative of $\sin(x^3)$ can be derived by expanding $\sin(x^3)$ into its Maclaurin series and then integrating term by term.

Step-by-step solution:

1. Recall the Maclaurin series for $\sin(x)$: $\sin(x) = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} x^{2n+1}.$

2. Substitute $x^3$ into the series for $\sin(x)$: $\sin(x^3) = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} (x^3)^{2n+1}.$

   Simplify the exponents: $\sin(x^3) = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} x^{3(2n+1)} = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} x^{6n+3}.$

3. Antidifferentiate term by term: To find the antiderivative, integrate each term: $\int \sin(x^3) \, dx = \int \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} x^{6n+3} \, dx.$

   Integrating term by term: $\int x^{6n+3} \, dx = \frac{x^{6n+4}}{6n+4}.$

   Thus, the series for the antiderivative becomes: $\int \sin(x^3) \, dx = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!(6n+4)} x^{6n+4} + C,$ where $C$ is the constant of integration.

Final Answer:

The Maclaurin series for the antiderivative of $\sin(x^3)$ is: $\int \sin(x^3) \, dx = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!(6n+4)} x^{6n+4} + C.$

Would you like further clarifications or a specific numerical example?

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Here are 5 related questions you might find interesting:

1. How would the Maclaurin series change if it were for $\cos(x^3)$ instead?
2. Can this series be approximated numerically for a specific value of $x$?
3. What is the radius of convergence of this series?
4. How does the term-by-term integration work for other functions, like $e^{x^3}$?
5. Could this series be truncated for a simpler approximation?

Tip: When dealing with integrals of non-elementary functions, series expansions are often the best tool for approximations and further analysis!