91Ó°ÊÓ

Hyperbolic functions are analogs of trigonometric functions. They are important in calculus due to their unique properties and applications. Two primary hyperbolic functions are the hyperbolic sine ( \(\sinh(u)\) ) and hyperbolic cosine ( \(\cosh(u)\) ).

• The hyperbolic sine function, \(\sinh(u)\) , is defined by \(\sinh(u) = \frac{e^u - e^{-u}}{2}\) .
• The hyperbolic cosine function, \(\cosh(u)\) , is defined as \(\cosh(u) = \frac{e^u + e^{-u}}{2}\) .

Hyperbolic functions describe the shape of a hanging cable or chain, known as a catenary, in physics. They have similar properties to trigonometric functions: they have derivatives and integrals that resemble those found in trigonometry. For instance, \(\frac{d}{du} [\sinh(u)] = \cosh(u)\) and \(\frac{d}{du} [\cosh(u)] = \sinh(u)\).
Recognizing these relationships helps us solve integrals and differential equations involving hyperbolic functions.

Differentiation is a fundamental tool in calculus which involves finding the rate at which a function is changing at any given point. When solving integrals involving substitution, it's necessary to determine the derivative of the function being substituted. In this exercise, we dealt with the differentiation of \(u = x^{-1/2}\) .To differentiate, we use the power rule which states that the derivative of \(x^n\) is \(nx^{n-1}\) . Applying this rule to our substitution:

• \( u = x^{-1/2} \) indicates \( \frac{du}{dx} = -\frac{1}{2} x^{-3/2} \) .

Understanding how to differentiate the substitution helps in transforming the integral into a form that is easier to evaluate. The relationship between the substitution and its derivative is essential for determining the differential in terms of \(du\).

Calculus integrals allow us to find areas under curves, central to many problems in mathematics and physics. There are various techniques for integration, one being \(u\) -substitution, which simplifies integrals by changing variables.In the given exercise, \(u\) -substitution is used to integrate the function \(\sinh(x^{-1/2}) / x^{3/2}\). This method involves:

• Identifying a substitution that reduces complexity, such as \(u = x^{-1/2}\).
• Using \(du\) to express \(dx\) and adjust the integral.

Once simplified, the integral becomes more straightforward to evaluate. In this case, integrating \(\sinh(u)\) produces \(-2 \cosh(u) \).
Subsequently, back-substituting \(u\) with \(x^{-1/2}\) provides the antiderivative of the original expression. Mastery of \(u\) -substitution is crucial to efficiently solving integrals, especially when complicated functions involve hyperbolic terms.