91Ó°ÊÓ

Hyperbolic functions, similar to trigonometric functions, arise in many areas of mathematics and physics. They are based on exponential functions and have properties that resemble those of sine and cosine functions. The main hyperbolic functions are the hyperbolic sine, \( \sinh(x) \), and the hyperbolic cosine, \( \cosh(x) \). These are defined as follows:

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

When dealing with hyperbolic functions, there are several identities that can simplify problems, such as \( \sinh^2(x) = \frac{1 - \cosh(2x)}{2} \). This particular identity is useful in calculus when simplifying integrals involving these functions. In the context of our problem, using the identity \( \sinh^2\left(\frac{x}{2}\right) = \frac{1 - \cosh(x)}{2} \) helps to transform the problem into a more manageable form.

Integration techniques involve various methods to solve integrals, which are fundamental in calculus. In our example, the strategy was to apply a hyperbolic identity to simplify the original integral. This falls under the technique known as substitution, where a transformation is made to streamline the equation.

Upon substituting the hyperbolic identity, the integral becomes simpler. It transforms into an integral we can solve with basic methods, such as the integration of constant values or simple functions.

• For constant integrals, like \( \int k \, dx = kx \), simply multiply the constant by the variable.
• For hyperbolic functions, familiar rules such as \( \int \cosh(x) \, dx = \sinh(x) + C \) can be applied.

The key here is to break down challenging expressions using identities or transformations, which often reveals simpler integrals that are easier to solve.

Definite integrals are used to determine the net area under a curve over a specified interval. They produce a finite number and are denoted by limits on their integral sign, such as \( \int_{a}^{b} f(x) \, dx \). In the problem at hand, \( \int_{0}^{\ln 10} 4 \sinh^{2}\left(\frac{x}{2}\right) \, dx \), becomes a definite integral problem once the limits of integration are established.

After simplifying the integral expression using hyperbolic identities, the integration limits from 0 to \( \ln 10 \) determine the range over which the area is calculated. Solving the integral involves evaluating the antiderivative at the upper limit and subtracting the value at the lower limit. In practice, this reflects on how an integral can provide quantitative measurements, such as calculating areas or quantities over a given interval.