To express the integral $\int_{-1}^{1} \frac{1}{\sqrt{1 - x^8}} \, dx$ in terms of the Beta function $B(x,y)$, we first note the relation between the Beta function and certain types of integrals.

The Beta function $B(x,y)$ is defined as: $B(x, y) = \int_{0}^{1} t^{x-1} (1 - t)^{y-1} \, dt$

For our integral, we need to manipulate it into a form that resembles the Beta function definition.

Step 1: Symmetry and Simplification

As previously noted, the integrand $\frac{1}{\sqrt{1 - x^8}}$ is an even function, so we can rewrite the integral as: $\int_{-1}^{1} \frac{1}{\sqrt{1 - x^8}} \, dx = 2 \int_{0}^{1} \frac{1}{\sqrt{1 - x^8}} \, dx$

Step 2: Variable Substitution

To relate the integral to the Beta function, we use the substitution $x^8 = t$, which gives $dx = \frac{1}{8} t^{-\frac{7}{8}} dt$. The limits of integration change as follows:

• When $x = 0$, $t = 0$
• When $x = 1$, $t = 1$

Thus, the integral becomes: $\int_{0}^{1} \frac{1}{\sqrt{1 - x^8}} \, dx = \int_{0}^{1} \frac{1}{\sqrt{1 - t}} \cdot \frac{1}{8} t^{-\frac{7}{8}} \, dt$

Step 3: Expressing the Integral in Beta Function Form

Rewriting the integrand, we get: $\int_{0}^{1} \frac{1}{8} t^{-\frac{7}{8}} (1 - t)^{-\frac{1}{2}} \, dt$

This integral now resembles the Beta function form: $\int_{0}^{1} t^{x-1} (1 - t)^{y-1} \, dt$

By comparing the exponents, we identify $x - 1 = -\frac{7}{8}$ and $y - 1 = -\frac{1}{2}$, leading to $x = \frac{1}{8}$ and $y = \frac{1}{2}$. Therefore: $\int_{0}^{1} \frac{1}{\sqrt{1 - x^8}} \, dx = \frac{1}{8} B\left(\frac{1}{8}, \frac{1}{2}\right)$

Step 4: Doubling the Result for the Original Integral

Given that: $\int_{-1}^{1} \frac{1}{\sqrt{1 - x^8}} \, dx = 2 \int_{0}^{1} \frac{1}{\sqrt{1 - x^8}} \, dx$

We have: $\int_{-1}^{1} \frac{1}{\sqrt{1 - x^8}} \, dx = 2 \cdot \frac{1}{8} B\left(\frac{1}{8}, \frac{1}{2}\right) = \frac{1}{4} B\left(\frac{1}{8}, \frac{1}{2}\right)$

Final Result

Thus, the integral $\int_{-1}^{1} \frac{1}{\sqrt{1 - x^8}} \, dx$ expressed in terms of the Beta function is: $\int_{-1}^{1} \frac{1}{\sqrt{1 - x^8}} \, dx = \frac{1}{4} B\left(\frac{1}{8}, \frac{1}{2}\right)$

Would you like more details or have any questions?

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Here are some related questions for further exploration:

1. How do you compute the Beta function $B(x, y)$ for specific values of $x$ and $y$?
2. What are some common integral forms that can be expressed using the Beta function?
3. How is the Beta function related to the Gamma function?
4. What are some applications of the Beta function in probability and statistics?
5. Can you express the integral in terms of the Gamma function using the relationship between Beta and Gamma functions?
6. How do you evaluate integrals involving square roots of polynomial expressions?
7. What numerical methods are available for evaluating special functions like the Beta function?
8. What are some other special functions that arise in integral calculus?

Tip: Understanding the relationships between special functions like the Beta and Gamma functions can greatly simplify the evaluation of complex integrals.