The integral in the image is:

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

To solve this integral, let's proceed with the following steps:

Step 1: Consider the substitution

Let: $u = \sin^{-1}(x)$ This implies: $\sin(u) = x$ Differentiating both sides with respect to $x$, we get: $\cos(u) \, du = dx \quad \text{and} \quad \cos(u) = \sqrt{1 - \sin^2(u)} = \sqrt{1 - x^2}$

Thus, $dx = \frac{du}{\cos(u)}$ and $\cos(u) = \sqrt{1 - x^2}$.

Step 2: Rewrite the integral in terms of $u$

Substitute into the integral: $\int \frac{\sin^{-1}(x)}{\sqrt{1 - x^2}} \, dx = \int \frac{u}{\cos(u)} \cdot \cos(u) \, du$ Simplifying, we get: $\int u \, du$

Step 3: Integrate with respect to $u$

The integral of $u$ with respect to $u$ is straightforward: $\int u \, du = \frac{u^2}{2} + C$ Where $C$ is the constant of integration.

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

Finally, substitute back $u = \sin^{-1}(x)$ to get: $\frac{(\sin^{-1}(x))^2}{2} + C$

Final Answer

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

Would you like further details or have any other questions?

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Here are some related questions you might explore:

1. What is the derivative of the inverse sine function $\sin^{-1}(x)$?
2. How can substitution be used to solve other integrals involving inverse trigonometric functions?
3. How do you evaluate the definite integral of this function from 0 to 1?
4. What are the integrals involving inverse cosine $\cos^{-1}(x)$ or inverse tangent $\tan^{-1}(x)$?
5. How do we interpret the integral $\int \frac{1}{\sqrt{1 - x^2}} \, dx$ geometrically?
6. What role do inverse trigonometric functions play in calculus and physics?
7. How does integration by parts compare to the method of substitution for this type of integral?
8. Can you generalize this integral to involve higher powers of the inverse sine function?

Tip: When solving integrals involving inverse trigonometric functions, substitution is often a powerful method, particularly when combined with recognizing the derivative of the inverse function itself.