The expression you've provided looks like the integral of the square of the arcsine function. The correct interpretation seems to be:

$\int (\arcsin(x))^2 \, dx$

This is the integral of $(\arcsin(x))^2$ with respect to $x$.

To solve this, we can proceed with integration by parts. Let's walk through the steps.

Steps:

1. Substitute variables for integration by parts: Let: $u = (\arcsin(x))^2$ $dv = dx$

2. Differentiate $u$ and integrate $dv$: $du = 2\arcsin(x) \cdot \frac{1}{\sqrt{1 - x^2}} \, dx$ (since the derivative of $\arcsin(x)$ is $\frac{1}{\sqrt{1 - x^2}}$)

   $v = x$

3. Apply the integration by parts formula: The integration by parts formula is: $\int u \, dv = uv - \int v \, du$

   Substituting the values for $u$, $du$, $v$, and $dv$: $\int (\arcsin(x))^2 \, dx = x (\arcsin(x))^2 - \int x \cdot \left(2 \arcsin(x) \cdot \frac{1}{\sqrt{1 - x^2}} \right) \, dx$

4. Simplify the remaining integral: The remaining integral is: $2 \int x \cdot \frac{\arcsin(x)}{\sqrt{1 - x^2}} \, dx$ This integral can be solved using substitution or known integral tables.

Final Answer (after simplification):

The solution to this integral is:

$x (\arcsin(x))^2 - 2 \arcsin(x) + 2\sqrt{1 - x^2} + C$

where $C$ is the constant of integration.

Would you like more details on any step? Or have any specific questions?

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Here are 5 related questions for practice:

1. What is the integral of $\arcsin(x)$?
2. How would you approach the integral of $(\arccos(x))^2$?
3. What is the derivative of $\arcsin(x)$?
4. How do you apply integration by parts in general?
5. Can you evaluate $\int (\arctan(x))^2 \, dx$?

Tip: Always consider substitution when dealing with integrals involving inverse trigonometric functions.