To evaluate the integral $I = \int_0^1 \frac{dx}{\sqrt{1 - x^4}}$ we can use the Beta function, which is related to the Gamma function.

The Beta function $B(x,y)$ is defined as: $B(x,y) = \int_0^1 t^{x-1} (1-t)^{y-1} \, dt$

We will rewrite the given integral in a form that can be expressed using the Beta function. First, let's make a substitution: $x^4 = t$ $4x^3 \, dx = dt$ $dx = \frac{dt}{4x^3} = \frac{dt}{4t^{3/4}}$

Next, adjust the limits of integration according to the substitution:

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

Substituting these into the integral, we get: $I = \int_0^1 \frac{1}{\sqrt{1 - t}} \cdot \frac{dt}{4t^{3/4}} = \frac{1}{4} \int_0^1 \frac{t^{-3/4}}{\sqrt{1 - t}} \, dt$

This integral can be recognized as a Beta function: $\frac{1}{4} \int_0^1 t^{-3/4} (1-t)^{-1/2} \, dt = \frac{1}{4} B\left(\frac{1}{4}, \frac{1}{2}\right)$

The Beta function can be related to the Gamma function: $B(x,y) = \frac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)}$

Therefore: $B\left(\frac{1}{4}, \frac{1}{2}\right) = \frac{\Gamma\left(\frac{1}{4}\right) \Gamma\left(\frac{1}{2}\right)}{\Gamma\left(\frac{1}{4} + \frac{1}{2}\right)} = \frac{\Gamma\left(\frac{1}{4}\right) \Gamma\left(\frac{1}{2}\right)}{\Gamma\left(\frac{3}{4}\right)}$

We know that $\Gamma\left(\frac{1}{2}\right) = \sqrt{\pi}$ and use the reflection formula for the Gamma function: $\Gamma(x) \Gamma(1-x) = \frac{\pi}{\sin(\pi x)}$ So: $\Gamma\left(\frac{1}{4}\right) \Gamma\left(\frac{3}{4}\right) = \frac{\pi}{\sin\left(\frac{3\pi}{4}\right)} = \frac{\pi}{\frac{\sqrt{2}}{2}} = \pi \sqrt{2}$ Thus: $\Gamma\left(\frac{3}{4}\right) = \frac{\pi \sqrt{2}}{\Gamma\left(\frac{1}{4}\right)}$

Plugging this into the Beta function expression: $B\left(\frac{1}{4}, \frac{1}{2}\right) = \frac{\Gamma\left(\frac{1}{4}\right) \sqrt{\pi}}{\frac{\pi \sqrt{2}}{\Gamma\left(\frac{1}{4}\right)}} = \frac{\Gamma\left(\frac{1}{4}\right)^2 \sqrt{\pi}}{\pi \sqrt{2}}$

Finally: $I = \frac{1}{4} B\left(\frac{1}{4}, \frac{1}{2}\right) = \frac{1}{4} \cdot \frac{\Gamma\left(\frac{1}{4}\right)^2 \sqrt{\pi}}{\pi \sqrt{2}}$

Simplifying this: $I = \frac{\Gamma\left(\frac{1}{4}\right)^2 \sqrt{\pi}}{4 \pi \sqrt{2}}$

Hence: $I = \frac{\Gamma\left(\frac{1}{4}\right)^2 \sqrt{\pi}}{4 \pi \sqrt{2}}$

Would you like to go through any of the details or have any questions on this?

Here are five questions you might ask:

1. How do the Gamma and Beta functions relate to each other?
2. What is the reflection formula for the Gamma function?
3. How can we simplify expressions involving Gamma functions?
4. Can you explain the substitution method used in solving integrals?
5. What are some applications of the Gamma and Beta functions in mathematics?

Tip: When dealing with integrals, transforming the integral into a known form using substitutions can simplify the evaluation process significantly.