Problem 37 - Thomas Calculus in SI Units

Question

Verify the integration formulas. a. \(\int \operatorname{sech} x d x=\tan ^{-1}(\sinh x)+C\) b. \(\int \operatorname{sech} x d x=\sin ^{-1}(\tanh x)+C\)

Step-by-Step Solution

Answer

Both integration formulas are correct.

Step 1: Recall the Definitions of Sech, Sinh, and Tanh

The hyperbolic secant function is defined as \( \operatorname{sech} x = \frac{1}{\cosh x} \). The hyperbolic sine function is \( \sinh x = \frac{e^x - e^{-x}}{2} \), and the hyperbolic tangent function is \( \tanh x = \frac{\sinh x}{\cosh x} \).

Step 2: Differentiate the Right Side of Formula (a)

To verify the formula \( \int \operatorname{sech} x \, dx = \tan^{-1}(\sinh x) + C \), differentiate \( \tan^{-1}(\sinh x) \). Use the chain rule: \( \frac{d}{dx}[\tan^{-1}(u)] = \frac{1}{1+u^2} \cdot \frac{du}{dx} \) with \( u = \sinh x \). Since \( \frac{d}{dx}(\sinh x) = \cosh x \), we have \( \frac{1}{1+(\sinh x)^2} \cdot \cosh x \).

Step 3: Simplify the Derivative from Step 2

The expression \( 1 + (\sinh x)^2 = \cosh^2 x \), so the derivative becomes \( \frac{\cosh x}{\cosh^2 x} = \operatorname{sech} x \). This confirms that differentiating \( \tan^{-1}(\sinh x) \) yields \( \operatorname{sech} x \), verifying formula (a).

Step 4: Differentiate the Right Side of Formula (b)

To verify the formula \( \int \operatorname{sech} x \, dx = \sin^{-1}(\tanh x) + C \), differentiate \( \sin^{-1}(\tanh x) \). Use the chain rule: \( \frac{d}{dx}[\sin^{-1}(u)] = \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx} \), with \( u = \tanh x \). Since \( \frac{d}{dx}(\tanh x) = \operatorname{sech}^2 x \), we have \( \frac{\operatorname{sech}^2 x}{\sqrt{1-(\tanh x)^2}} \).

Step 5: Simplify the Derivative from Step 4

The expression \( 1 - (\tanh x)^2 = \operatorname{sech}^2 x \). Thus, the derivative becomes \( \frac{\operatorname{sech}^2 x}{\sqrt{\operatorname{sech}^2 x}} = \operatorname{sech} x \). This confirms that differentiating \( \sin^{-1}(\tanh x) \) also yields \( \operatorname{sech} x \), verifying formula (b).

Key Concepts

Hyperbolic Functions

Hyperbolic functions are similar to trigonometric functions, but they are defined using exponentials rather than angles. The important hyperbolic functions involved in this exercise are:

• Hyperbolic Secant (\(\operatorname{sech} x\)): Defined as the reciprocal of the hyperbolic cosine, it is given by \(\operatorname{sech} x = \frac{1}{\cosh x}\).
• Hyperbolic Sine (\(\sinh x\)): Often used in modeling hyperbolas, it is expressed as \(\sinh x = \frac{e^x - e^{-x}}{2}\).
• Hyperbolic Tangent (\(\tanh x\)): The ratio of hyperbolic sine to hyperbolic cosine, \(\tanh x = \frac{\sinh x}{\cosh x}\).

These functions model various natural phenomena, like the shape of cables hanging under their own weight. They are interconnected through identities that are derived from their exponential definitions, much like their circular trigonometric counterparts.

Integration Formulas

Integration is the reverse operation of differentiation. It finds the original function from its derivative. The integration formulas addressed here are:

• \(\int \operatorname{sech} x \, dx = \tan^{-1}(\sinh x) + C\)
• \(\int \operatorname{sech} x \, dx = \sin^{-1}(\tanh x) + C\)

To verify these formulas, one must differentiate the right side of each equation to check if it returns the left side.
It is important to note that both expressions deal with hyperbolic integrals involving \(\operatorname{sech}\), and finding such integrals requires using methods involving trigonometric identities or substitutions not immediately obvious from basic rules.

Differentiation

Differentiation involves calculating the derivative of a function, which represents the rate of change of the function with respect to a variable. In the provided exercise, differentiation is used to verify integration formulas.
The derivatives of inverse trigonometric and hyperbolic functions play a critical role:

• For \(\tan^{-1}(\sinh x)\), the derivative involves calculating \(\frac{d}{dx}\left[\tan^{-1}(u)\right] = \frac{1}{1+u^2} \cdot \frac{du}{dx}\).
• For \(\sin^{-1}(\tanh x)\), the formula \(\frac{d}{dx}\left[\sin^{-1}(u)\right] = \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx}\) is used.

Understanding and applying these formulas help decipher the solutions in the integration formulas.

Chain Rule

The chain rule is a fundamental concept in calculus used to differentiate composite functions. It's crucial for this exercise when differentiating inverse hyperbolic functions.
When you have a function nested inside another, the rule states:

• \(\frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x)\)

This means you take the derivative of the outer function and multiply it by the derivative of the inner function.
In Step 2 of verifying formula (a), for instance, \(\frac{d}{dx} [\tan^{-1}(\sinh x)]\) involves differentiating \(\tan^{-1}\) and multiplying by the derivative of \(\sinh x\), which is \(\cosh x\).
For formula (b), applying the chain rule to \(\sin^{-1}(\tanh x)\) involves using the formula for \(\sin^{-1}\) and multiplying it by \(\operatorname{sech}^2 x\), the derivative of \(\tanh x\).
By practicing the chain rule, it becomes easier to handle complex differentiations as seen in these examples.