Question

Carry out the following steps to derive the formula \(\int \operatorname{csch} x d x=\ln |\tanh (x / 2)|+C\) (Theorem 9). a. Change variables with the substitution \(u=x / 2\) to show that $$\int \operatorname{csch} x d x=\int \frac{2 d u}{\sinh 2 u}$$. b. Use the identity for \(\sinh 2 u\) to show that \(\frac{2}{\sinh 2 u}=\frac{\operatorname{sech}^{2} u}{\tanh u}\). c. Change variables again to determine \(\int \frac{\operatorname{sech}^{2} u}{\tanh u} d u,\) and then express your answer in terms of \(x\).

Short answer

#Answer# Based on the given step by step solution, the short answer for the integral can be written as: $\int \operatorname{csch} x dx = \ln |\tanh (x / 2)|+C$.

Step-by-step solution

Substitute Variables

Let's first substitute \(u = x/2\). Then, \(du = \frac{1}{2} dx\), and \(dx = 2 du\). Now, rewrite the integral: $$\int \operatorname{csch} x dx = \int \frac{1}{\operatorname{sinh} x} dx = \int \frac{2 du}{\operatorname{sinh} 2u}$$.

Use Identity for \(\operatorname{sinh} 2u\)

We're given the identity: \(\operatorname{sinh} 2u = 2\operatorname{sinh} u\operatorname{cosh} u\). Using this, we can rewrite the expression: $$\frac{2}{\operatorname{sinh} 2u} = \frac{1}{\operatorname{sinh} u\operatorname{cosh} u} = \frac{\operatorname{sech}^2 u}{\tanh u}$$.

Change Variables Again

Let's make another substitution: \(v = \tanh u\). Then, \(dv = \operatorname{sech}^2 u du\). Our integral now becomes: $$\int \frac{\operatorname{sech}^2 u}{\tanh u} du = \int \frac{dv}{v}$$. Evaluate this integral: $$\int \frac{dv}{v} = \ln |v| + C = \ln |\tanh u| + C$$. Now substitute back in terms of \(x\): $$\ln |\tanh u| + C = \ln |\tanh (\frac{x}{2})| + C$$. So, we have derived the formula: $$\int \operatorname{csch} x dx=\ln |\tanh (x / 2)|+C$$.

Key concepts

Hyperbolic Functions

Hyperbolic functions are analogs of trigonometric functions but for hyperbolas instead of circles. The most common hyperbolic functions are \ \(\sinh\ \) (hyperbolic sine), \ \(\cosh\ \) (hyperbolic cosine), \(\tanh\ \) (hyperbolic tangent), and \(\operatorname{csch}\ \) (hyperbolic cosecant). These functions often appear in calculus and differential equations due to their many useful identities and properties.

Some key hyperbolic function identities include:

• The identity for hyperbolic sine: \(\sinh 2u = 2\sinh u \cosh u\)
• The relationship between hyperbolic sine and cosecant: \(\operatorname{csch} x = \frac{1}{\sinh x}\)

In integration techniques, hyperbolic functions are treated similarly to trigonometric functions. They are highly useful in substitutions and solving certain types of integrals. Understanding their identities helps simplify and solve complex integrals, just like in the example problem where \(\operatorname{csch} x\) was manipulated using hyperbolic identities to find its integral.

Substitution Method

The substitution method is a powerful technique for solving integrals. It involves replacing a part of the integral with a new variable to simplify the expression. This is particularly useful when dealing with complex integrands.

For the given problem, we start with the substitution \(u = \frac{x}{2}\). This changes the variable from \(x\) to \(u\), making it easier to handle the integral. As a result, the differential \(dx\) is transformed to \(2 du\), thus altering the entire integral's structure:

• Original integral: \(\int \operatorname{csch} x\, dx\)
• Transformed integral: \(\int \frac{2 du}{\sinh 2u}\)

By substituting wisely, we simplify complex expressions, allowing us to use known identities and solve them with greater ease. The key point is understanding how the substitution affects each part of the integral, which lets us unravel more complicated functions into manageable pieces.

Integral Identities

Integral identities can vastly simplify the process of integration by converting complicated expressions into more familiar forms. These identities are derived from fundamental equations and are immensely valuable in calculus.
In the example problem, the identity for \(\sinh 2u = 2\sinh u \cosh u\) was used to simplify the integrand. Recognizing integral identities allows us to decompose expressions and make further substitutions or simplifications. The integration process often depends on:

• Recognizing a pattern or identity that matches part of the integral.
• Applying the identity to simplify the expression.

In the problem at hand, the use of the identity for \(\frac{2}{\sinh 2u}\) as \(\frac{\operatorname{sech}^2 u}{\tanh u}\) was essential. This transformation led directly to a simpler integral, \(\int \frac{dv}{v}\), whose solution is the natural logarithm, \(\ln |v|\). This step reinforced how knowing and applying integral identities can streamline finding antiderivatives.