Question

Assume that \(f\) has an inverse on its domain. a. Let \(y=f^{-1}(x),\) which means \(x=f(y)\) and \(d x=f^{\prime}(y) d y\) Show that $$\int f^{-1}(x) d x=\int y f^{\prime}(y) d y.$$ b. Use the result of Exercise 65 to show that $$\int f^{-1}(x) d x=y f(y)-\int f(y) d y.$$ c. Use the result of part (b) to evaluate \(\int \ln x d x\) (express the result in terms of \(x\) ). d. Use the result of part (b) to evaluate \(\int \sin ^{-1} x d x\). e. Use the result of part (b) to evaluate \(\int \tan ^{-1} x d x\).

Short answer

b. What is the relationship between the integral of \(f^{-1}(x)\) and the integral of \(f(y)\)? c. What is the integral of \(\ln x\) with respect to x? d. What is the integral of \(\sin^{-1} x\) with respect to x? e. What is the integral of \(\tan^{-1} x\) with respect to x? Answers: a. The substitution method is used to prove the equation. b. The relationship is: \(\int f^{-1}(x) dx = yf(y)-\int f(y) dy\) c. The integral of \(\ln x\) with respect to x is: \(x\ln x - x\) d. The integral of \(\sin^{-1} x\) with respect to x is: \(x\sin^{-1}x+\int \sqrt{1-x^2} dx\) e. The integral of \(\tan^{-1} x\) with respect to x is: \(x\tan^{-1}x-\int \frac{x}{1+x^2} dx\)

Step-by-step solution

Use substitution method

We can make the substitution \(x=f(y)\), so \(dx=f^{\prime}(y)dy\). Therefore, the integral changes to: $$\int f^{-1}(x) dx = \int y f^{\prime}(y) dy$$ #b. Applying the result of Exercise 65#

Apply Exercise 65

Using the result of Exercise 65, we have an equation of the form $$\int u(x)v(x)dx=u(x)v(x)-\int u^{\prime}(x)v(x)dx.$$ In our case, let \(u(x)=y\) and \(v(x)=f^{\prime}(y)\), so that: $$u^{\prime}(x)=1 \Longrightarrow\int y f^{\prime}(y) dy=yf(y)-\int f(y) dy$$ Therefore, we have: $$\int f^{-1}(x) dx = yf(y)-\int f(y) dy$$ #c. Evaluating \(\int \ln x dx\)#

Use the result from part (b)

We have the result from part (b): \(\int f^{-1}(x) dx=yf(y)-\int f(y) dy\). Since we are given that \(f^{-1}(x)=\ln x\), we can use the fact that \(f(x)=e^x\) and calculate the integral: $$\int \ln x dx = x\ln x-x$$ #d. Evaluating \(\int \sin^{-1} x dx\)#

Use the result from part (b)

We have the result from part (b): \(\int f^{-1}(x) dx=yf(y)-\int f(y) dy\). Since we are given that \(f^{-1}(x)=\sin^{-1}x\), we can use the fact that \(f(x)=\sin x\) and calculate the integral: $$\int \sin^{-1} x dx = x\sin^{-1}x+\int \sqrt{1-x^2} dx$$ #e. Evaluating \(\int \tan^{-1} x dx\)#

Use the result from part (b)

We have the result from part (b): \(\int f^{-1}(x) dx=yf(y)-\int f(y) dy\). Since we are given that \(f^{-1}(x)=\tan^{-1}x\), we can use the fact that \(f(x)=\tan x\) and calculate the integral: $$\int \tan^{-1} x dx = x\tan^{-1}x-\int \frac{x}{1+x^2} dx$$

Key concepts

Inverse Functions

Inverse functions reverse the effect of a given function. If we have a function \(f\) such that \(f(y) = x\), then the inverse, denoted \(f^{-1}\), undoes this operation. So, if you apply \(f^{-1}\) to \(x\), you get back \(y\) such that \(f(y) = x\).

When dealing with integration, integrating the inverse function \(f^{-1}\) with respect to \(x\) is the same as integrating \(y\) with respect to \(f(x)\). This concept plays a crucial role in understanding how to solve integrals involving inverse functions, as shown in part (b) of the exercise. Understanding the properties of inverse functions and their derivatives is key to simplifying complex integrals.

Substitution Method

The substitution method is a technique used to simplify the process of integration by changing variables. In integration problems, especially when dealing with inverse functions, substitution can make the integral more manageable.

In this exercise, the method involves letting \(x=f(y)\), and subsequently, the differential \(dx\) becomes \(f^{\prime}(y) dy\). This transforms the original integral \(\int f^{-1}(x) dx\) into \(\int y f^{\prime}(y) dy\), which is often easier to evaluate. This is because the new integral is expressed in terms of \(y\) and its derivative, conforming to a familiar integration form.

Integral Calculus

Integral calculus is a fundamental branch of mathematics focused on the accumulation of quantities and areas under and between curves. It includes techniques for finding integrals and antiderivatives.

The exercise demonstrates the integration of inverse functions, a concept rooted deeply in integral calculus. Understanding how to apply integration techniques, like the substitution method, reveals how functions accumulate or transform across intervals. Immersing in integral calculus communicates how to

• find solutions to integral equations,
• evaluate the area under curves, and
• determine accumulated quantity over a region.

This core concept is essential for grasping the complete evaluation of functions, such as logarithmic and trigonometric integrals, which are presented in further steps of the solution.

Trigonometric Integrals

Trigonometric integrals involve the integration of trigonometric functions, such as \(\sin(x)\), \(\cos(x)\), and their inverses. These integrals often require specialized techniques due to the unique properties of trigonometric functions.

In this exercise, the integrals \(\int \sin^{-1}(x) dx\) and \(\int \tan^{-1}(x) dx\) are evaluated using the integration principle derived from inverse functions. Knowing how to deal with inverse trigonometric functions involves:

• Understanding the geometric interpretation of these functions,
• Applying integration by parts or substitution methods, and
• Recognizing trigonometric identities that can simplify the process.

By mastering these principles, students can better evaluate more complex trigonometric integrals, leveraging known derivatives and integrals to simplify calculations.