∫_(-1)^1▒〖[arcsin⁡(u) ]^2 ⅆu〗

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.

Learn how to solve the definite integral ∫_{-1}^{1} [arcsin(u)]^2 du by using symmetry of even functions and integration by parts. This problem covers key calculus concepts, including inverse trigonometric functions, symmetry, and advanced integration techniques. Suitable for undergraduate students in Calculus II or III.

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