37–56. Integrals Evaluate each integral.
∫ sinh x / (1 + cosh x) dx

Verified step by step guidance

1. Recall the definitions and identities for hyperbolic functions: \(\sinh x = \frac{e^x - e^{-x}}{2}\) and \(\cosh x = \frac{e^x + e^{-x}}{2}\). Also, remember the identity \$1 + \cosh x = 2 \cosh^2 \left(\frac{x}{2}\right)$.
2. Rewrite the integral using the identity for the denominator: \(\int \frac{\sinh x}{1 + \cosh x} \, dx = \int \frac{\sinh x}{2 \cosh^2 \left(\frac{x}{2}\right)} \, dx\).
3. Express \(\sinh x\) in terms of \(\sinh \frac{x}{2}\) and \(\cosh \frac{x}{2}\) using the double-angle formula: \(\sinh x = 2 \sinh \left(\frac{x}{2}\right) \cosh \left(\frac{x}{2}\right)\).
4. Substitute this expression into the integral to get \(\int \frac{2 \sinh \left(\frac{x}{2}\right) \cosh \left(\frac{x}{2}\right)}{2 \cosh^2 \left(\frac{x}{2}\right)} \, dx\), which simplifies to \(\int \frac{\sinh \left(\frac{x}{2}\right)}{\cosh \left(\frac{x}{2}\right)} \, dx\).
5. Recognize that \(\frac{\sinh u}{\cosh u} = \tanh u\), where \(u = \frac{x}{2}\). Use substitution \(u = \frac{x}{2}\), so \(dx = 2 du\), and rewrite the integral as \(\int \tanh u \cdot 2 \, du = 2 \int \tanh u \, du\). Then, integrate \(\tanh u\) using its known antiderivative.

Key Concepts

Hyperbolic Functions

Hyperbolic functions, such as sinh x and cosh x, are analogs of trigonometric functions but based on hyperbolas. They have specific identities like cosh²x - sinh²x = 1, which are useful for simplifying expressions and integrals involving these functions.

Integration Techniques for Rational Functions

When integrating a ratio of functions, it is often helpful to simplify the integrand by substitution or algebraic manipulation. Recognizing patterns or rewriting the integrand in terms of a single variable can make the integral more straightforward to solve.

Substitution Method

The substitution method involves changing variables to simplify an integral. By letting u equal a function inside the integral, you can rewrite the integral in terms of u and du, making it easier to integrate, especially when the derivative of u appears elsewhere in the integrand.