Question

For the following exercises, find the antiderivatives for the given functions.\(\frac{\sinh (x)}{1+\cosh (x)}\)

Short answer

The antiderivative is \( 2 \ln\left|\cosh\left(\frac{x}{2}\right)\right| + C \).

Step-by-step solution

Recognize the Function Form

The function given is \( \frac{\sinh(x)}{1+\cosh(x)} \). This suggests a hyperbolic trigonometric expression, which often can be simplified using identities.

Use Hyperbolic Identity

Recall the hyperbolic identity: \( \cosh^2(x) - \sinh^2(x) = 1 \). This identity can be used to simplify hyperbolic expressions. Also, note the identity \( 1 + \cosh(x) = 2\cdot \cosh^2\left(\frac{x}{2}\right) \).

Simplify the Expression

Re-write the denominator using the identity: \( 1 + \cosh(x) = 2\cdot \cosh^2\left(\frac{x}{2}\right) \). The expression becomes:\[ \frac{\sinh(x)}{2\cdot \cosh^2\left(\frac{x}{2}\right)} \]Using the identity \( \sinh(x) = 2\cdot \sinh\left(\frac{x}{2}\right)\cosh\left(\frac{x}{2}\right) \).Substitute to get:\[ \frac{2\cdot \sinh\left(\frac{x}{2}\right)\cosh\left(\frac{x}{2}\right)}{2\cdot \cosh^2\left(\frac{x}{2}\right)} = \frac{\sinh\left(\frac{x}{2}\right)}{\cosh\left(\frac{x}{2}\right)} \].

Further Simplification to Hyperbolic Tangent

Recognize that the fraction \( \frac{\sinh\left(\frac{x}{2}\right)}{\cosh\left(\frac{x}{2}\right)} \) simplifies to \( \tanh\left(\frac{x}{2}\right) \).

Integrate the Simplified Function

Find the antiderivative of \( \tanh\left(\frac{x}{2}\right) \). The integral is:\[ \int \tanh\left(\frac{x}{2}\right) \, dx \]Use the substitution \( u = \frac{x}{2} \), which gives \( du = \frac{1}{2} dx \) or \( dx = 2 \, du \).

Change of Variables in the Integral

With the change of variables:\[ \int \tanh(u) \, 2 \, du = 2 \int \tanh(u) \, du \]The antiderivative of \( \tanh(u) \) is \( \ln(|\cosh(u)|) \).

Apply the Antiderivative Formula

The integral becomes:\[ 2 \cdot \ln(|\cosh(u)|) + C \]Substitute back \( u = \frac{x}{2} \):\[ 2 \cdot \ln\left(\left|\cosh\left(\frac{x}{2}\right)\right|\right) + C \].

Key concepts

Hyperbolic Functions

Hyperbolic functions are essential in calculus, similar to trigonometric functions, but they involve exponential functions. They are defined using the exponential functions \(e^x\) and \(e^{-x}\). The most common hyperbolic functions include the hyperbolic sine \(\sinh(x)\) and the hyperbolic cosine \(\cosh(x)\). These are defined as:

• \(\sinh(x) = \frac{e^x - e^{-x}}{2}\)
• \(\cosh(x) = \frac{e^x + e^{-x}}{2}\)

Hyperbolic functions have many properties in parallels with trigonometric functions, like hyperbolic identities. An example is the identity \(\cosh^2(x) - \sinh^2(x) = 1\), similar to the Pythagorean identity. These functions are notably useful in solving integrals that involve their form. Recognizing and manipulating these functions through identities and simplifications guide us to find the antiderivative effectively.
Understanding these functions and their properties ensures you can simplify expressions and tackle a wide array of calculus problems efficiently.

Trigonometric Identities

Trigonometric identities aren't just critical in trigonometry but extend their utility to hyperbolic functions, which have similar identities. Understanding them allows us to simplify expressions, making integration much easier. For hyperbolic functions, analogous identities exist, such as:

• \(\cosh^2(x) - \sinh^2(x) = 1\)
• \(1 + \cosh(x) = 2\cdot \cosh^2\left(\frac{x}{2}\right)\)

The original problem uses these identities to rewrite the expression \(\frac{\sinh(x)}{1+\cosh(x)}\). By expressing \(1 + \cosh(x)\) using \(2\cdot \cosh^2\left(\frac{x}{2}\right)\), we simplify the problem significantly.
Key identities are crucial tools in calculus to switch seemingly complex problems into handleable forms. They not only simplify computations but also uncover deeper connections between different mathematical functions.

Integration Techniques

Integration techniques encompass various methods scientists use to find antiderivatives. Sometimes, the integration of a function isn't straightforward and requires transformation techniques. For instance, we can use substitution when integrating the function \(\tanh\left(\frac{x}{2}\right)\). The substitution method involves setting a part of the expression equal to \(u\), simplifying the integral.In the given solution, the problem involves using:

• Substitution: Let \(u = \frac{x}{2}\), then \(dx = 2 \, du\).

This substitution transforms the integral into a more solvable form. Recognition of such techniques is essential, particularly when faced with complex functions like hyperbolic expressions.
Generalized integration approaches, like substitution and recognizing patterns, are core skills in calculus. They not only help solve a vast range of integral problems but also reveal the connectivity between various mathematical ideas.