Problem 40 - Thomas Calculus

Question

Verify the integration formulas in Exercises \(37-40\). $$ \int \tanh ^{-1} x d x=x \tanh ^{-1} x+\frac{1}{2} \ln \left(1-x^{2}\right)+C $$

Step-by-Step Solution

Answer

Integration is verified as differentiation matches the integrand.

Step 1: Differentiate the Right Side

To verify the integration formula, differentiate the right side of the equation: \[ \frac{d}{dx} \left(x \tanh^{-1} x + \frac{1}{2} \ln (1 - x^2) + C \right). \] Differentiate each term individually.

Step 2: Differentiate the First Term

The first term is \( x \tanh^{-1} x \). Use the product rule for differentiation: \[ \frac{d}{dx}(u \, v) = u' \, v + u \, v', \] where \( u = x \) and \( v = \tanh^{-1} x \). Calculate \( u' = 1 \) and \( v' = \frac{1}{1-x^2} \). Therefore, \[ \frac{d}{dx}(x \tanh^{-1} x) = \tanh^{-1} x + x \cdot \frac{1}{1-x^2}. \]

Step 3: Differentiate the Second Term

Consider the second term \( \frac{1}{2} \ln (1-x^2) \). Use the chain rule for differentiation: \[ \frac{d}{dx} \ln(g(x)) = \frac{1}{g(x)} g'(x), \] where \( g(x) = 1-x^2 \). Compute \( g'(x) = -2x \). Thus, \[ \frac{d}{dx} \left( \frac{1}{2} \ln (1-x^2) \right) = \frac{1}{2} \cdot \frac{-2x}{1-x^2} = \frac{-x}{1-x^2}. \]

Step 4: Combine the Results

Combine the derivatives obtained from Steps 2 and 3:\[ \tanh^{-1} x + x \cdot \frac{1}{1-x^2} - \frac{x}{1-x^2}. \]Simplify the expression:\[ \tanh^{-1} x + \frac{x}{1-x^2} - \frac{x}{1-x^2} = \tanh^{-1} x. \]

Step 5: Compare with the Left Side

Since the derivative of the right side, \( \tanh^{-1} x \), matches the given integrand, it verifies that:\[ \int \tanh^{-1} x \, dx = x \tanh^{-1} x + \frac{1}{2} \ln(1-x^2) + C. \]

Key Concepts

Inverse Hyperbolic Functions

Inverse hyperbolic functions are analogs of the standard inverse trigonometric functions. They relate to hyperbolic functions such as sinh, cosh, and tanh. These functions have important applications in calculus, particularly in integration, as they provide unique integrals and derivatives that differ from trigonometric functions. The inverse hyperbolic tangent, written as \( \tanh^{-1} x \), is of particular interest. It represents the value whose hyperbolic tangent is \( x \). This function is defined for \( |x| < 1 \) and has a derivative of \( \frac{1}{1-x^2} \).

Recognizing and understanding inverse hyperbolic functions is essential for solving integrals involving them, especially since they appear in many physical models and mathematical equations. Using these functions, one can often simplify complex integrals into more manageable expressions.

Integration Techniques

Integration techniques are methods used to find the integrals of functions. In calculus, finding integrals is essential because it provides the area under a curve, among other things. The integration example in the exercise used several techniques: the product rule and the chain rule. Recognizing which techniques to use is critical to solving integration problems efficiently.

When the integrand is complex, like \( \tanh^{-1} x \), splitting it into several parts allows for different rules to be applied individually. Sometimes it involves algebraic manipulation, using substitution, or integrating by parts. Being adept at selecting appropriate integration techniques simplifies the process and ensures accuracy in the results.

Product Rule

The product rule is a fundamental rule in calculus used for differentiating expressions involving products of two functions. If \( u(x) \) and \( v(x) \) are two functions, then their product's derivative is given by:

• \( (uv)' = u'v + uv' \)

The product rule was applied to differentiate the term \( x \tanh^{-1} x \) in the exercise. Here, \( u = x \) and \( v = \tanh^{-1} x \), leading to the derivatives \( u' = 1 \) and \( v' = \frac{1}{1-x^2} \). Applying the rule helps to separate the terms into more straightforward expressions.

The product rule is essential for handling mixed terms, especially when the functions involved have their own derivatives, and those derivatives are relatively easy to compute. Understanding when and how to apply the product rule streamlines calculus tasks.

Chain Rule

The chain rule is a crucial technique in calculus used to differentiate composite functions. When you have a function of another function, the chain rule enables you to compute the derivative. If \( y = f(g(x)) \), the chain derivative is given by:

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

In the context of the given exercise, the chain rule was used for the term \( \ln(1 - x^2) \). Here, \( g(x) = 1 - x^2 \) with the derivative \( g'(x) = -2x \). By applying the chain rule, we discovered that the derivative is \( \frac{-x}{1-x^2} \).

The chain rule is invaluable for tackling complex derivatives that involve nested functions. Knowing how to use the chain rule broadens your ability to handle a variety of problems in calculus, facilitating deeper understanding and application.