Exercises 71 and 72 compare the results of using Equations...

Exercises 71 and 72 compare the results of using Equations $(4)$ and $(5)$ Equations $(4)$ and $(5)$ give different formulas for the integral of $\cos ^{-1} x :$ a. $\int \cos ^{-1} x d x=x \cos ^{-1} x-\sin \left(\cos ^{-1} x\right)+C$ b. $\int \cos ^{-1} x d x=x \cos ^{-1} x-\sqrt{1-x^{2}}+C$ Can both integrations be correct? Explain.

Exercises 71 and 72 compare the results of using Equations $(4)$ and $(5)$
Equations $(4)$ and $(5)$ give different formulas for the integral of $\cos ^{-1} x :$
a. $\int \cos ^{-1} x d x=x \cos ^{-1} x-\sin \left(\cos ^{-1} x\right)+C$
b. $\int \cos ^{-1} x d x=x \cos ^{-1} x-\sqrt{1-x^{2}}+C$
Can both integrations be correct? Explain.

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Key Concepts

Antiderivatives and the Integration Constant

In indefinite integration, any two antiderivatives of a function are considered equivalent if they differ only by a constant. This is because the derivative of a constant is zero, making the constant arbitrary. Therefore, seemingly different integration results can actually represent the same family of functions when the constant of integration is taken into account.

Integration of Inverse Trigonometric Functions

Integrating inverse trigonometric functions, such as the inverse cosine function, often involves techniques like substitution or integration by parts. The process may yield expressions that look different based on the method used, but these expressions can be equivalent when algebraically simplified or when rearranged to include a differing constant.

Trigonometric Identities Involving Inverse Functions

Trigonometric identities provide a connection between the function and its inverse. For example, the identity sin(arccos(x)) = ?(1-x²) shows how expressions using sine of an inverse cosine can be rewritten in terms of x. This identity explains why two integration results, one including sin(arccos(x)) and the other ?(1-x²), can be equivalent up to a constant.

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