Problem 41 - Essential Calculus

Question

Find the integral. \(\int \frac{\cosh x}{\sinh x} d x\)

Step-by-Step Solution

Answer

\(\ln|\sinh(x)| + C\

Step 1: Identify the type of integral

This is not a basic integral, and might need substitution. The denominator \(\sinh(x)\) is the derivative of the numerator \(\cosh(x)\), so substitution will be wise to proceed.

Step 2: Apply substitution

Let \(u = \sinh(x)\). The derivative of \(\sinh(x)\) is \(\cosh(x)\), so \(du = \cosh(x) dx\). Now, substituting in the integral, it becomes: \(\int \frac{1}{u} du\).

Step 3: Solve the integral

Now that the integral is simplified, we can solve it. The integral of \(\frac{1}{u}\) with respect to \(u\) is \(\ln |u|\). Therefore, the solution to our integral is \(\ln |u|\).

Step 4: Substitute back for \(u\)

Remember from step 2, \(u\) was defined as \(\sinh(x)\). Substituting this back in gives the final solution as \(\ln|\sinh(x)|\). However we should not forget the constant of integration \(C\), therefore the final answer is \(\ln|\sinh(x)| + C\)

Key Concepts

Integration by Substitution

Understanding integration by substitution, sometimes known as u-substitution, is critical for solving complex integrals where the standard rules do not apply directly. The basic idea is to simplify the integral by substituting a part of the original function with a new variable, usually denoted as u. This technique often reveals a simpler integral that we can solve using basic integration rules.

To apply this method, we need to identify a portion of the integrand (the function being integrated) that has a derivative elsewhere in the integrand. Then we can set this portion equal to u and compute the differential du. In our example, we let u be equal to \(\sinh(x)\), which simplifies the integral remarkably, as its derivative \(\cosh(x)\) is present in the numerator. The process transforms the integral into one in terms of u, making the integral more manageable.

Hyperbolic Functions

The hyperbolic functions are analogs of the trigonometric functions but for a hyperbola, as opposed to a circle. They include functions like \(\sinh(x)\), \(\cosh(x)\), and \(\tanh(x)\), among others, and have similar properties to trigonometric functions but behave differently in terms of their graphs and identities.

In integration, these functions often pose unique challenges. For example, the connection between \(\sinh(x)\) and \(\cosh(x)\) is analogous to that of sine and cosine in trigonometry, with \(\cosh(x)\) being the derivative of \(\sinh(x)\). This relationship is useful when applying techniques like substitution. As in our exercise, knowing that the derivative of \(\sinh(x)\) is \(\cosh(x)\) allows us to execute a well-placed substitution, simplifying the integral significantly.

Definite and Indefinite Integrals

When we look at definite and indefinite integrals, we are addressing two related but distinct concepts. An indefinite integral represents a family of functions and includes an arbitrary constant, typically denoted by \(C\). This constant represents the idea that when we differentiate a function, we lose information about vertical shifts.

On the other hand, a definite integral computes the net area under the curve of the function between two specific points on the x-axis and does not include a constant of integration. The problem we are examining presents an indefinite integral, as indicated by the absence of limits of integration. Thus, we include the + \(C\) after integrating, acknowledging the infinite family of antiderivatives associated with the given integrand.

Natural Logarithm as an Integral

The natural logarithm function, denoted \(\ln(x)\), is of particular interest in integration because it is the antiderivative of \(1/x\). This relationship is fundamental and often arises in calculus after making an appropriate substitution. When computing integrals like \(\int \frac{1}{u} du\), we end up with \(\ln|u|\), plus a constant if it's an indefinite integral.

In our example, after substituting \(u = \sinh(x)\), we find that the integral of \(1/u\) leads us directly to \(\ln|u|\). It's a powerful tool, and recognizing when it can be applied simplifies many complex integrals enormously. The absolute value in the logarithm function accounts for the domain of \(\ln(x)\) since the natural logarithm is only defined for positive arguments.