Problem 29 - Calculus

Question

Find antiderivatives for the functions in \(25-32\) \(\frac{\sinh x}{1+\cosh x}\)

Step-by-Step Solution

Answer

The antiderivative is \( \frac{1}{2}x + \frac{1}{2}e^{-x} + C \).

Step 1: Identify the Integral

We need to find the antiderivative of the function \( \frac{\sinh x}{1 + \cosh x} \). This involves integrating the function with respect to \( x \).

Step 2: Simplify the Integrand

Notice that \( 1 + \cosh x = e^x \). Therefore, the integrand \( \frac{\sinh x}{1 + \cosh x} \) can be rewritten as \( \frac{\sinh x}{e^x} \).

Step 3: Rewrite \(\sinh x\) in Exponential Form

Recall that \( \sinh x = \frac{e^x - e^{-x}}{2} \). Substituting this in, we have: \[ \frac{\sinh x}{e^x} = \frac{\frac{e^x - e^{-x}}{2}}{e^x} = \frac{e^x - e^{-x}}{2e^x} = \frac{1}{2} - \frac{e^{-x}}{2} \].

Step 4: Integrate Term by Term

Now, integrate each term separately:For \( \frac{1}{2} \):\[ \int \frac{1}{2} \, dx = \frac{1}{2} x \].For \( -\frac{e^{-x}}{2} \):\[ \int -\frac{e^{-x}}{2} \, dx = \frac{1}{2} e^{-x} \].

Step 5: Combine Results and Add Constant of Integration

Putting the results from Step 4 together, the antiderivative is:\[ \frac{1}{2}x + \frac{1}{2}e^{-x} + C \]where \( C \) is the constant of integration.

Key Concepts

Integration Techniques

When dealing with integrals, choosing the right integration technique can significantly simplify the process. In our case, we're finding the antiderivative of the function \( \frac{\sinh x}{1 + \cosh x} \). The first step is recognizing that integration often involves transforming or rewriting complex expressions. Here, that means expressing the integrand in a form that is easier to integrate with respect to \( x \).

Begin by transforming the integrand. Noticing that \( 1 + \cosh x = e^x \) is key, as it allows us to rewrite the function as \( \frac{\sinh x}{e^x} \). With the use of hyperbolic identities, we transform \( \sinh x = \frac{e^x - e^{-x}}{2} \), helping us simplify to \( \frac{1}{2} - \frac{e^{-x}}{2} \).

This decomposition makes it easier: each term can now be integrated straightforwardly. Such transformations are part of an important set of integration techniques that rely on logical algebraic manipulations and familiar transformations. With a complex function, always keep an eye out for potential substitutions or simplifications.

Hyperbolic Functions

Hyperbolic functions are analogs of trigonometric functions that arise often in calculations involving exponential growth or hyperbolic geometry. Two central hyperbolic functions are \( \sinh x \) and \( \cosh x \), defined as:

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

These functions have properties and identities similar to those of trigonometric functions, such as \( \cosh^2 x - \sinh^2 x = 1 \).

In the given problem, understanding these identities allows for simplification by rewriting \( 1 + \cosh x = e^x \). These expressions form an integral part of simplifying the integrand due to their ties to exponential functions. Recognizing these relationships opens up the path to more straightforward integration techniques!

Constant of Integration

One important aspect of integrating functions is the constant of integration, denoted as \( C \). Whenever you compute an indefinite integral, it’s essential to include this \( C \) in your final answer.

So, why is this so significant? When you take the antiderivative of a function, what you are actually finding is a family of functions. Each function in this family differs by a constant. Consider a simple case of integrating a constant function, leading to a linearly increasing function, each differing by a vertical shift.
By including the constant of integration, you're addressing the fact that there are infinitely many functions that have the same derivative of your original function, but differ by that constant amount. In our problem, after combining the results, the final antiderivative is \( \frac{1}{2}x + \frac{1}{2}e^{-x} + C \), ensuring that all possible vertical translations of the antiderivative function are captured!