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?

---

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.

Explore the detailed steps to solve the integral of (arcsin(x))^2 using integration by parts. This guide explains each step, from substitution to simplification, with related formulas and concepts. Ideal for undergraduate students studying calculus.

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