91Ó°ÊÓ

Integration by parts leads to a rule for integrating inverses that usually gives good results: $$ \begin{aligned} \int f^{-1}(x) d x &=\int y f^{\prime}(y) d y \\ &=y f(y)-\int f(y) d y \\ &=x f^{-1}(x)-\int f(y) d y \end{aligned} $$ The idea is to take the most complicated part of the integral, in this case \(f^{-1}(x),\) and simplify it first. For the integral of ln \(x,\) we get $$ \begin{aligned} \int \ln x d x &=\int y e^{y} d y \\ &=y e^{y}-e^{y}+C \\\ &=x \ln x-x+C \end{aligned} $$ For the integral of \(\cos ^{-1} x\) we get $$ \begin{aligned} \int \cos ^{-1} x d x &=x \cos ^{-1} x-\int \cos y d y \\ &=x \cos ^{-1} x-\sin y+C \\ &=x \cos ^{-1} x-\sin \left(\cos ^{-1} x\right)+C \end{aligned} $$ Use the formula $$ \int f^{-1}(x) d x=x f^{-1}(x)-\int f(y) d y $$ to evaluate the integrals in Exercises \(71-74 .\) Express your answers in terms of \(x .\) $$ \int \sec ^{-1} x d x $$

Step by step solution

01

Setup for Integration by Parts Formula

To evaluate the integral \( \int \sec^{-1}(x) \, dx \), start by applying the integration by parts formula \( \int f^{-1}(x) \, dx = x f^{-1}(x) - \int f(y) \, dy \). Here, \( f^{-1}(x) = \sec^{-1}(x) \). The goal is to identify \( f(x) \) such that \( f^{-1}(x) = \sec^{-1}(x) \).

02

Recognize Function for Secant Inverse

Consider the function \( f(x) = \sec(x) \). Thus, \( f^{-1}(x) = \sec^{-1}(x) \) implies that \( y = \sec^{-1}(x) \) and \( f(y) = \sec(y) \). We then apply these functions in the integration by parts formula.

03

Apply the Integration by Parts Formula

Use the identified function \( f(y) = \sec(y) \) in the formula. Therefore, \( \int \sec^{-1}(x) \, dx = x \sec^{-1}(x) - \int \sec(y) \, dy \).

04

Evaluate Integral of Function

Compute \( \int \sec(y) \, dy \). The antiderivative of \( \sec(y) \) is known to be \( \ln |\sec(y) + \tan(y)| \). Therefore, substitution into the integration by parts formula gives \( x \sec^{-1}(x) - \ln |\sec(y) + \tan(y)| + C \).

05

Substitute Back In Terms of x

Remembering that \( y = \sec^{-1}(x) \) means \( \sec(y) = x \) and \( \tan(y) = \sqrt{x^2 - 1} \). Substitute back to get the final expression: \( x \sec^{-1}(x) - \ln |x + \sqrt{x^2 - 1}| + C \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Inverse Functions

Inverse functions reverse the operation of a function. If you have a function \( f(x) \), the inverse function, denoted as \( f^{-1}(x) \), will undo the operation of \( f(x) \). For example, if \( f(x) = \sec(x) \), then \( f^{-1}(x) = \sec^{-1}(x) \), meaning that \( \sec^{-1}(x) \) will give you the angle whose secant is \( x \). Understanding inverse functions is crucial for solving integrals involving inverse trigonometric functions, like \( \sec^{-1}(x) \), which require applying integration techniques that consider these properties. They allow us to switch between functions and their inverses, playing a key role when setting up problems with integration by parts.

Antiderivatives

An antiderivative of a function \( f(x) \) is a function whose derivative is \( f(x) \). If \( F'(x) = f(x) \), then \( F(x) \) is an antiderivative of \( f(x) \). This concept is central in calculus because finding antiderivatives allows us to evaluate definite and indefinite integrals.
For example, in the given problem, the antiderivative of \( \sec(y) \), known to be \( \ln |\sec(y) + \tan(y)| \), is used in the integration by parts formula. Learning how to find antiderivatives of functions, especially trigonometric and inverse trigonometric functions, helps evaluate complicated integrals efficiently.

Integration Techniques

Integration techniques are methods used to determine the antiderivative or integral of a function. One essential technique is integration by parts, which is especially useful when dealing with products of functions or inverse trigonometric functions. The integration by parts formula is given by\[\int u \, dv = uv - \int v \, du\]where \( u \) and \( dv \) are parts of the integrand we choose to differentiate and integrate, respectively.

• For example, in the integral \( \int \sec^{-1}(x) \, dx \), integration by parts involves identifying appropriate \( u \) and \( dv \) that match the form \( f^{-1}(x) \) as given in the problem.
• After choosing \( u = \sec^{-1}(x) \) and \( dv = dx \), the formula transforms the integral into simpler parts that can further be resolved using known antiderivatives.

By mastering these techniques, especially integration by parts, tackling integrals involving inverse functions becomes much easier, resulting in elegant and simplified expressions.