To solve the integral

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

using an appropriate substitution, we can approach it by simplifying the components, particularly by using a substitution involving the inverse sine function.

Step-by-Step Solution:

1. Choose a Substitution: Let's set $u = \sin^{-1}(x)$. Then, $x = \sin(u)$.

2. Differentiate to Find $dx$: By differentiating $x = \sin(u)$, we get $dx = \cos(u) \, du.$

3. Substitute $x$ and $dx$ in the Integral: Notice that $\sqrt{1 - x^2} = \sqrt{1 - \sin^2(u)} = \cos(u)$. Now, substitute these expressions into the integral: $\int \frac{2}{\sqrt{1 - x^2} \sin^{-1}(x)} \, dx = \int \frac{2}{\cos(u) \cdot u} \cdot \cos(u) \, du.$

4. Simplify the Integral: The $\cos(u)$ terms cancel out, leaving us with $\int \frac{2}{u} \, du.$

5. Integrate: Now, integrate with respect to $u$: $\int \frac{2}{u} \, du = 2 \ln |u| + C.$

6. Back-Substitute $u = \sin^{-1}(x)$: Replace $u$ with $\sin^{-1}(x)$ to get the final answer: $2 \ln |\sin^{-1}(x)| + C.$

Final Answer:

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

Would you like further explanation on any of these steps or additional practice problems?

Here are some related questions you may find helpful:

1. What substitution would work for integrals involving $\tan^{-1}(x)$ instead of $\sin^{-1}(x)$?
2. How would you solve an integral with a different inverse trigonometric function, like $\cos^{-1}(x)$?
3. What techniques are useful when integrals contain both square roots and trigonometric functions?
4. How does the chain rule apply to inverse trigonometric functions during differentiation?
5. Can you solve this integral by a different method, such as integration by parts?

Tip: For integrals involving inverse trigonometric functions, substitutions that simplify the inverse function (like $u = \sin^{-1}(x)$) can be particularly effective for simplifying the expression.