Question

Integrals Involving Hyperbolic Functions Evaluate the following integrals: a. \(\int x \cosh \left(x^{2}\right) d x\) b. \(\int \tanh x d x\)

Short answer

a. \( \frac{1}{2} \sinh(x^2) + C \) b. \( \ln|\cosh(x)| + C \)

Step-by-step solution

Identify the Integral Type for Part (a)

The integral \( \int x \cosh(x^2) \, dx \) suggests that substitution could simplify the integration process. In particular, notice the argument \( x^2 \) inside the hyperbolic cosine, which hints at using a substitution related to \( x^2 \).

Execute Substitution for Part (a)

Let \( u = x^2 \), then \( du = 2x \, dx \), or \( x \, dx = \frac{1}{2} \, du \). Substitute these into the integral to get:\[ \int x \cosh(x^2) \, dx = \int \frac{1}{2} \cosh(u) \, du. \]

Integrate with Respect to the New Variable for Part (a)

The integral becomes:\[ \frac{1}{2} \int \cosh(u) \, du. \]The antiderivative of \( \cosh(u) \) is \( \sinh(u) \). So, we have:\[ \frac{1}{2} \sinh(u) + C. \]

Substitute Back to Original Variables for Part (a)

Replace \( u \) with \( x^2 \) to express the solution in terms of \( x \):\[ \frac{1}{2} \sinh(x^2) + C. \]

Identify the Integral Type for Part (b)

The integral \( \int \tanh(x) \, dx \) is a well-known standard integral. We aim to find an antiderivative for \( \tanh(x) \).

Use Known Antiderivative for Part (b)

The antiderivative of \( \tanh(x) \) is \( \ln|\cosh(x)| \). Thus, the integral becomes:\[ \ln|\cosh(x)| + C. \]

Key concepts

Hyperbolic Function Integration

When integrating hyperbolic functions, the approach is similar to integrating trigonometric functions, since hyperbolic functions share similar properties. Hyperbolic functions like \( anh(x)\), \( ext{sech}(x)\), and \( ext{csch}(x)\) have standard integrals much like \( an(x)\) or \( ext{sec}(x)\) in the realm of trigonometry. Analyzing the given exercise, we are expected to evaluate two integrals involving hyperbolic functions:

• \(\int x \cosh(x^2) \, dx\)
• \(\int \tanh x \, dx\)

Hyperbolic function integration can sometimes be simplified by recognizing parts of the function that can be substituted. Just like with trigonometric integrals, knowledge of these standard forms and their antiderivatives is crucial. However, unlike trigonometric functions, hyperbolic functions do not oscillate, which sometimes makes their integration smoother.

Substitution Method

The substitution method in calculus is a powerful tool used when a direct integration approach is difficult or cumbersome. In the context of hyperbolic functions, substitution is particularly valuable, as seen in the first integral exercise \( \int x \cosh(x^2) \, dx \). To simplify this integral, we use substitution:

• Let \(u = x^2\), which implies that \(du = 2x \, dx\). This rearranges to \(x \, dx = \frac{1}{2} \, du\).

Then, substitute these into the integral:\[\int x \cosh(x^2) \, dx = \int \frac{1}{2} \cosh(u) \, du\]This transformation allows us to work with a simpler hyperbolic integral, \(\int \cosh(u) \, du \), which is straightforward to integrate. This method not only streamlines the integration process but also turns a seemingly complex problem into one that is manageable.

Antiderivative of Hyperbolic Functions

Understanding antiderivatives of hyperbolic functions is essential for solving integrals involving these functions. Hyperbolic functions, much like their trigonometric counterparts, have specific antiderivatives that need to be memorized for efficient problem-solving. For the exercise's problems:

• The integral \(\int \cosh(u) \, du\) leads to its antiderivative \(\sinh(u)\). Therefore, after substitution and integration, and reverting to the original variables, we have the expression \(\frac{1}{2} \sinh(x^2) + C \).
• For the integral \(\int \tanh(x) \, dx\), recognizing the form is the key. The antiderivative of \(\tanh(x)\) is \(\ln|\cosh(x)|\). This is directly applied to solve the second part of the exercise, providing the solution as \(\ln|\cosh(x)| + C\).

In both cases, the knowledge of these basic antiderivatives facilitates the integration process and helps avoid unnecessary complexity. Remembering these forms and practicing their application in various contexts is an essential skill in calculus.