Let's solve the integral using the substitution method explicitly, as requested:

The integral we need to solve is:

$\int \frac{\sin^{-1}(x)}{\sqrt{1 - x^2}} \, dx$

Step 1: Substitute the inverse sine function

Let: $u = \sin^{-1}(x)$ This substitution gives: $\sin(u) = x$ Differentiating both sides with respect to $x$: $\cos(u) \, du = dx$ where $\cos(u) = \sqrt{1 - \sin^2(u)} = \sqrt{1 - x^2}$.

Step 2: Substitute into the integral

Now, substitute $u = \sin^{-1}(x)$ and $dx = \cos(u) \, du$ into the integral:

$\int \frac{\sin^{-1}(x)}{\sqrt{1 - x^2}} \, dx = \int \frac{u}{\sqrt{1 - \sin^2(u)}} \cdot \cos(u) \, du$

Since $\sqrt{1 - \sin^2(u)} = \cos(u)$, the integral simplifies to:

$\int \frac{u \cdot \cos(u)}{\cos(u)} \, du = \int u \, du$

Step 3: Integrate with respect to $u$

The integral is now straightforward:

$\int u \, du = \frac{u^2}{2} + C$

Step 4: Substitute back $u = \sin^{-1}(x)$

Finally, substitute back $u = \sin^{-1}(x)$ to express the integral in terms of $x$:

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

Final Answer

Thus, using the substitution method, we have:

$\frac{(\sin^{-1}(x))^2}{2} + C$

This matches the result obtained earlier. Would you like further clarification or additional assistance with a related problem?

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Here are some related questions you might consider:

1. What are the general steps for solving integrals using substitution?
2. How does substitution work differently for other inverse trigonometric functions like $\cos^{-1}(x)$ or $\tan^{-1}(x)$?
3. What other integrals can be solved using a similar substitution involving trigonometric identities?
4. Can this integral be solved using integration by parts instead of substitution?
5. How would you verify the result of this integral by differentiating the answer?
6. What is the geometric interpretation of the integral $\int \frac{1}{\sqrt{1 - x^2}} \, dx$?
7. How would you evaluate this integral if the limits of integration were given, say from 0 to 1?
8. What are some applications of inverse trigonometric functions in physics or engineering?

Tip: Always check the derivative of your result to ensure correctness after solving an integral, especially when using substitution methods.