Question

There are several ways to express the indefinite integral of sech \(x\). a. Show that \(\int \operatorname{sech} x d x=\tan ^{-1}(\sinh x)+C\) (Theorem 9 ). (Hint: Write sech \(x=\frac{1}{\cosh x}=\frac{\cosh x}{\cosh ^{2} x}=\frac{\cosh x}{1+\sinh ^{2} x},\) and then make a change of variables.) b. Show that \(\int \operatorname{sech} x d x=\sin ^{-1}(\tanh x)+C .\) (Hint: Show that sech \(x=\frac{\operatorname{sech}^{2} x}{\sqrt{1-\tanh ^{2} x}}\) and then make a change of variables.) c. Verify that \(\int \operatorname{sech} x d x=2 \tan ^{-1}\left(e^{x}\right)+C\) by proving \(\frac{d}{d x}\left(2 \tan ^{-1}\left(e^{x}\right)\right)=\operatorname{sech} x\).

Short answer

Question: Verify the three different equivalent expressions for the indefinite integral of \(\operatorname{sech} x\). Answer: We can verify the expressions by following these steps: a. \(\int \operatorname{sech} x dx = \tan^{-1}(\sinh x) + C\) can be shown by rewriting the integrand, making an appropriate substitution, and integrating. b. Similarly, \(\int \operatorname{sech} x dx = \sin^{-1}(\tanh x) + C\) can be verified by rewriting the integrand, making a suitable substitution, and integrating. c. To verify \(\int \operatorname{sech} x dx = 2 \tan^{-1}(e^x) + C\), we can differentiate the given function with respect to \(x\) and simplify the expression to get the \(\operatorname{sech} x\).

Step-by-step solution

a. Show that \(\int \operatorname{sech} x d x=\tan^{-1}(\sinh x)+C\)

First, rewrite the integrand as \(\frac{\cosh x}{1 + \sinh^2 x}\). Then, substitute \(u=\sinh x\), so \(du = \cosh x dx\). The integral now becomes: \(\int \frac{1}{1+u^2} du\) Now, integrate with respect to \(u\): \(\int \frac{1}{1+u^2} du = \tan^{-1}(u) + C = \tan^{-1}(\sinh x) + C\)

b. Show that \(\int \operatorname{sech} x d x=\sin ^{-1}(\tanh x)+C\)

Rewrite the integrand as \(\frac{\operatorname{sech}^2 x}{\sqrt{1-\tanh^2 x}}\). Then, substitute \(v = \tanh x\), so \(dv = \operatorname{sech}^2x \, dx\). The integral now becomes: \(\int \frac{1}{\sqrt{1-v^2}} dv\) Now, integrate with respect to \(v\): \(\int \frac{1}{\sqrt{1-v^2}} dv = \sin^{-1}(v) + C = \sin^{-1}(\tanh x) + C\)

c. Verify that \(\int \operatorname{sech} x d x=2 \tan^{-1}\left(e^{x}\right)+C\) by proving \(\frac{d}{d x}\left(2 \tan^{-1}\left(e^{x}\right)\right)=\operatorname{sech} x\)

First, differentiate the given function with respect to \(x\): \(\frac{d}{dx} \left[2 \tan^{-1} \left(e^x \right) \right] = 2 \frac{1}{1+(e^x)^2} \cdot \frac{d}{dx} \left(e^x \right)\) Now, simplify and evaluate the derivative: \(2 \frac{1}{1+(e^x)^2} \cdot e^x = 2 \frac{e^x}{1+e^{2x}}\) Rewrite the expression as: \(2 \frac{e^x}{\cosh^2 x} = 2 \frac{1}{2\cosh x} = \operatorname{sech} x\) Since the derivative of \(2 \tan^{-1}\left(e^{x}\right)\) is equal to \(\operatorname{sech} x\), we can conclude that \(\int \operatorname{sech} x d x=2 \tan ^{-1}\left(e^{x}\right)+C\).

Key concepts

Trigonometric Substitution

Trigonometric substitution is a powerful technique in integration. It helps simplify integrals that contain expressions like \( \sqrt{a^2 - x^2} \), \( \sqrt{a^2 + x^2} \), or \( \sqrt{x^2 - a^2} \). The core idea is to substitute a trigonometric function that transforms the expression into a simpler one, often using identities to replace the quadratic radical.

For example, consider the integral of \( \operatorname{sech} x \) from the exercise:

• The aim is to simplify \( \frac{\cosh x}{1 + \sinh^2 x} \).
• By substituting \( u = \sinh x \), the expression becomes \( \frac{1}{1+u^2} \), a form we can easily integrate using the inverse tangent formula.
• Thus, \( \int \frac{1}{1+u^2} \, du = \tan^{-1}(u) + C \), which leads to the solution \( \tan^{-1}(\sinh x) + C \).

These substitutions leverage the relationships between hyperbolic functions and their derivatives, making complex integrals manageable.

Hyperbolic Functions

Hyperbolic functions are analogs of trigonometric functions but for hyperbolas rather than circles. They include \( \sinh x \), \( \cosh x \), and \( \tanh x \), among others. These functions have properties and derivatives akin to their circular counterparts, which make them valuable in calculus, especially when dealing with hyperbolic substitutions.

Our exercise deals with converting the integral \( \int \operatorname{sech} x \, dx \) using hyperbolic identities:

• By expressing \( \operatorname{sech} x \) as \( \frac{\operatorname{sech}^2 x}{\sqrt{1 - \tanh^2 x}} \), we can use the variable substitution \( v = \tanh x \).
• This allows transforming the integral into \( \int \frac{1}{\sqrt{1 - v^2}} \, dv \), which is a format amenable to direct integration. The result is \( \sin^{-1}(v) + C \), or equivalently \( \sin^{-1}(\tanh x) + C \).

Understanding hyperbolic functions and their identities is crucial for simplifying and solving certain types of integrals.

Integration Techniques

Integration techniques are essential tools in calculus that help in evaluating integrals that do not have straightforward solutions. These techniques include substitution, integration by parts, partial fraction decomposition, and special function substitution.

In this exercise, several integration techniques come into play:

• Substitution: Used multiple times for rewriting the integrals in simpler forms, such as \( u = \sinh x \) or \( v = \tanh x \).
• Derivatives: Verification of the integral by differentiating functions as shown in the third part of the exercise. By proving the derivative of \( 2 \tan^{-1}(e^x) \) equals \( \operatorname{sech} x \), we verify, beyond integration, the correctness of our indefinite integral.

The efficient use of integration techniques transforms challenging problems into manageable tasks, enabling solutions that illuminate deeper mathematical connections.