1. The problem is to find the antiderivatives (indefinite integrals) of certain trigonometric-related functions. 2. The formulas for these antiderivatives are: $$\int \frac{dx}{\sqrt{1 - x^2}} = \sin^{-1} x + C$$ $$\int \frac{-dx}{\sqrt{1 - x^2}} = \cos^{-1} x + C$$ $$\int \frac{dx}{1 + x^2} = \tan^{-1} x + C$$ Note: The third integral in the user's message has a typo; it should be $\int \frac{dx}{1 + x^2}$ for $\tan^{-1} x$, not $1 - x^2$. 3. Explanation: - The integral of $\frac{1}{\sqrt{1 - x^2}}$ corresponds to the inverse sine function because the derivative of $\sin^{-1} x$ is $\frac{1}{\sqrt{1 - x^2}}$. - Similarly, the integral of $\frac{-1}{\sqrt{1 - x^2}}$ corresponds to the inverse cosine function. - The integral of $\frac{1}{1 + x^2}$ corresponds to the inverse tangent function. 4. These integrals are fundamental in calculus and are used to find antiderivatives involving inverse trigonometric functions. 5. Final answers: $$\int \frac{dx}{\sqrt{1 - x^2}} = \sin^{-1} x + C$$ $$\int \frac{-dx}{\sqrt{1 - x^2}} = \cos^{-1} x + C$$ $$\int \frac{dx}{1 + x^2} = \tan^{-1} x + C$$