The given integral is:

$I = \int_{-1}^{1} [\arcsin(u)]^2 \, du$

Symmetry of the function:

The function $[\arcsin(u)]^2$ is an even function because $\arcsin(-u) = -\arcsin(u)$, and squaring this makes it an even function. This simplifies the integral since the limits are symmetric around zero:

$I = 2 \int_{0}^{1} [\arcsin(u)]^2 \, du$

Now, let's solve this integral:

We can solve this integral using integration by parts.

Let:

• $v = [\arcsin(u)]^2$, so $dv = 2 \arcsin(u) \cdot \frac{1}{\sqrt{1 - u^2}} du$
• $du = du$

We'll also need to perform additional steps to fully evaluate this. Let me calculate the solution.

The value of the integral is:

$\int_{-1}^{1} [\arcsin(u)]^2 \, du = -4 + \frac{\pi^2}{2}$

Would you like more details or have any questions?

Here are five related questions you might find interesting:

1. What is the geometric interpretation of the $\arcsin(u)$ function?
2. How does the symmetry of a function affect definite integrals?
3. Can we use integration by parts to solve more complex trigonometric integrals?
4. How can we evaluate integrals with non-elementary functions like $\arcsin(u)$?
5. What are the properties of inverse trigonometric functions like $\arcsin(u)$?

Tip: When dealing with even functions and symmetric limits, you can often simplify the problem by integrating over half the interval and multiplying by 2.