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The integral provided is:

\[
\int_0^1 \frac{x^3}{\sqrt{1 - x^8}} \, dx
\]

### Approach:

We will attempt to solve this integral using a substitution method.

Let:
\[
u = x^4 \implies du = 4x^3 dx
\]

Now, adjust the limits of integration:
- When \( x = 0 \), \( u = 0 \)
- When \( x = 1 \), \( u = 1 \)

Now rewrite the integral in terms of \( u \):

\[
\int_0^1 \frac{x^3}{\sqrt{1 - x^8}} \, dx = \frac{1}{4} \int_0^1 \frac{du}{\sqrt{1 - u^2}}
\]

This is a standard integral, and it corresponds to the inverse sine function:

\[
\int \frac{du}{\sqrt{1 - u^2}} = \arcsin(u)
\]

Thus, the integral becomes:

\[
\frac{1}{4} \arcsin(u) \Big|_0^1
\]

Evaluating the limits:

\[
\frac{1}{4} \left( \arcsin(1) - \arcsin(0) \right) = \frac{1}{4} \left( \frac{\pi}{2} - 0 \right) = \frac{\pi}{8}
\]

### Final Answer:
\[
\int_0^1 \frac{x^3}{\sqrt{1 - x^8}} \, dx = \frac{\pi}{8}
\]

Would you like further details or have any questions? Here are five related questions:

1. How does substitution work in solving integrals?
2. What are other common trigonometric integrals that involve \(\arcsin\)?
3. Can substitution be applied to definite and indefinite integrals alike?
4. How is the integral of \(\frac{1}{\sqrt{1-u^2}}\) related to the unit circle?
5. Are there other approaches to solve integrals involving \( \sqrt{1 - x^8} \)?

**Tip:** For integrals involving \( \sqrt{1 - x^n} \), trigonometric substitution often simplifies the problem.

Evaluate the integral \( \int_0^1 \frac{x^3}{\sqrt{1 - x^8}} \, dx \)

Solve the integral \( \int_0^1 \frac{x^3}{\sqrt{1 - x^8}} \, dx \) using substitution and trigonometric identities. This problem covers advanced calculus concepts such as substitution in definite integrals and inverse trigonometric functions.

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