The given integral is:

$\int \frac{\arcsin x}{\sqrt{1 - x^2}} \, dx$

To solve this, we will use a substitution and some simplifications.

Step 1: Substitution

Let: $u = \arcsin x$ Then: $x = \sin u \quad \text{and} \quad dx = \cos u \, du$

Moreover, since: $\sqrt{1 - x^2} = \sqrt{1 - \sin^2 u} = \cos u,$ the integral becomes: $\int \frac{\arcsin x}{\sqrt{1 - x^2}} \, dx = \int \frac{u}{\cos u} \cdot \cos u \, du = \int u \, du$

Step 2: Integration of $u$

The integral of $u$ is straightforward: $\int u \, du = \frac{u^2}{2} + C$

Step 3: Back-Substitution

Recall that $u = \arcsin x$. Substituting back, we get: $\frac{u^2}{2} + C = \frac{(\arcsin x)^2}{2} + C$

Final Answer:

$\int \frac{\arcsin x}{\sqrt{1 - x^2}} \, dx = \frac{(\arcsin x)^2}{2} + C$

Let me know if you would like further clarification!

Related Questions:

1. How is the derivative of $\arcsin x$ derived?
2. What other substitution methods could be used for this integral?
3. Can this integral be solved using integration by parts?
4. How does the range of $\arcsin x$ influence the solution?
5. What would the solution look like if the limits of integration were given?

Tip:

When dealing with integrals involving $\arcsin x$, always consider substitution based on trigonometric identities to simplify the problem.