Question

In the following exercises, use the substitutions x=sinh?,cosh?, or tanh?. Express the final answers in terms of the variable x. $$ \int \frac{d x}{1-x^{2}} $$

Short answer

\( \tanh^{-1}(x) + C \)

Step-by-step solution

Choose the Appropriate Substitution

Since the integral is of the form \( \int \frac{dx}{1-x^2} \), it has a resemblance to the derivative of the inverse hyperbolic tangent function. Thus, we use the substitution \( x = \tanh(u) \). This means \( dx = \text{sech}^2(u) \, du \) and \( 1 - x^2 = \text{sech}^2(u) \).

Substitute and Simplify

Substitute \( x = \tanh(u) \) into the integral, which gives us \( \int \frac{\text{sech}^2(u) \, du}{\text{sech}^2(u)} = \int du \). The \( \text{sech}^2(u) \) terms cancel out.

Integrate and Find the Antiderivative

Solve the simplified integral, \( \int du \), which results in \( u + C \), where \( C \) is an integration constant.

Substitute Back Using the Original Variable

We have \( u = \tanh^{-1}(x) \), hence the solution of the integral in terms of \( x \) is \( \tanh^{-1}(x) + C \).

Key concepts

Understanding Hyperbolic Functions

Hyperbolic functions, much like their trigonometric counterparts, are deeply connected to the geometry of the hyperbola. Two of the most common hyperbolic functions are the hyperbolic sine, \( \sinh(x) \), and hyperbolic cosine, \( \cosh(x) \). However, in this exercise, we focus on \( \tanh(u) \), the hyperbolic tangent function. This function is defined as the ratio of \( \sinh(u) \) to \( \cosh(u) \), or \( \tanh(u) = \frac{\sinh(u)}{\cosh(u)} \). These functions have properties similar to trigonometric functions but relate to hyperbolas instead of circles.
Hyperbolic functions can be used effectively in calculus, particularly in integrals where expressions resemble hyperbolic identities or derivatives. Recognizing these forms can significantly simplify solving integrals.

Exploring Inverse Hyperbolic Functions

Inverse hyperbolic functions, like \( \tanh^{-1}(x) \), are useful when working backward from hyperbolic functions. They allow us to find the original angle \( u \) whose hyperbolic function would provide a certain value. The inverse hyperbolic tangent specifically, \( \tanh^{-1}(x) \), is used when \( x \) is set as \( \tanh(u) \). This function can conveniently help revert substitution steps in integration, making it possible to express the solution back in terms of the original variable. Thus, this step is essential for expressing complex integrals back in terms of the initial variables.

Applying the Substitution Method

The substitution method is a powerful technique in integration that simplifies problems by introducing a new variable. When given an integral, like \( \int \frac{dx}{1 - x^2} \), identifying a suitable substitution is crucial. Here, noticing that the expression resembles the derivative of \( \tanh^{-1}(x) \) hints at using the substitution \( x = \tanh(u) \). This choice simplifies the integral by exploiting the identities of hyperbolic functions.
After substitution, the integral \( \int \frac{\text{sech}^2(u) \, du}{\text{sech}^2(u)} \) easily reduces to \( \int du \). The simplicity after substitution not only makes the integration process easier but also demonstrates the problem-solving power of selecting appropriate substitution.

Finding the Antiderivative

The antiderivative is essentially the function whose derivative results in the original integrand. In integration, once we've simplified the integral to a form like \( \int du \), finding the antiderivative simply involves integrating \( \int du \), which yields \( u \). This result represents the general antiderivative of the expression.
It is crucial to reinstate the original variable after finding the antiderivative, which is where the substitution \( u = \tanh^{-1}(x) \) plays its role. Thus, the solution, \( \tanh^{-1}(x) + C \), is the antiderivative of the original expression in terms of \( x \).

Role of Integration Constants

Every indefinite integral includes an integration constant represented by \( C \). The constant exists because integration is the reverse process of differentiation, which does not account for constant terms. Consequently, integrating a function will always include \( C \) to represent any potential constant values that were present before differentiation.
It's vital not to forget to include this constant in the final solution, as it plays an important part in the complete expression representing all possible antiderivatives. So, when writing the integral of \( \int du \) as the final antiderivative, it is correctly expressed as \( u + C \), or \( \tanh^{-1}(x) + C \) after substitution back to the original variable.