Problem 41 - Thomas Calculus

Question

Evaluate the integrals in Exercises \(41-50\) . $$ \int \sinh 2 x d x $$

Step-by-Step Solution

Answer

\( \int \sinh 2x \, dx = \frac{1}{2} \cosh 2x + C \).

Step 1: Recall the Formula for Hyperbolic Functions

The integral we need to evaluate is \( \int \sinh 2x \, dx \). Recall that the hyperbolic sine function is defined as \( \sinh x = \frac{e^x - e^{-x}}{2} \). Its derivative is \( \cosh x \), and the derivative of \( \sinh 2x \) is \( 2 \cosh 2x \).

Step 2: Use Basic Integration Rule

The integral of \( \sinh ax \) is \( \frac{1}{a} \cosh ax + C \), where \( C \) is the constant of integration. In our case, \( a = 2 \).

Step 3: Apply the Integration Formula

Using the rule from Step 2, we integrate \( \sinh 2x \) to get \( \int \sinh 2x \, dx = \frac{1}{2} \cosh 2x + C \).

Step 4: Confirm the Result by Differentiation

To verify, differentiate \( \frac{1}{2} \cosh 2x + C \). The derivative \( \frac{1}{2} \cdot 2 \cosh 2x \) simplifies to \( \sinh 2x \), which matches the original integrand.

Key Concepts

Hyperbolic Functions

Hyperbolic functions are mathematical functions that are analogous to trigonometric functions. They are expressed using the exponential function. Two primary hyperbolic functions are hyperbolic sine (\(\sinh x\)) and hyperbolic cosine (\(\cosh x\)). The definition of the hyperbolic sine function is given by the expression: \[ \sinh x = \frac{e^x - e^{-x}}{2} \]. Similarly, the hyperbolic cosine is defined as \[ \cosh x = \frac{e^x + e^{-x}}{2} \].
These functions have derivatives similar to their trigonometric counterparts:

• The derivative of \(\sinh x\) is \(\cosh x\).
• The derivative of \(\cosh x\) is \(\sinh x\).

Understanding these derivatives is crucial.
It assists in calculating integrals involving hyperbolic functions, just as we do with trigonometric functions. For example, the integral of \(\sinh ax\) leads to \(\frac{1}{a}\cosh ax + C\), showcasing that hyperbolic integrals can often parallel those in trigonometry.

Definite Integral

In integral calculus, the definite integral represents the area under the curve of a function within a specific interval. It is calculated using the limit of a sum of areas of rectangles and is denoted by:\[ \int_{a}^{b} f(x) dx \].
The definite integral differs from the indefinite integral as it provides a numerical value, expressing a total between given bounds. In the context of hyperbolic functions, solving \(\int \sinh 2x \, dx\) could evolve into a definite integral such as \(\int_{0}^{0.5} \sinh 2x \, dx\).
To find this, follow these steps:

• Apply the integration formula: \(\int \sinh 2x \, dx = \frac{1}{2} \cosh 2x + C\).
• Evaluate the integral from \(0\) to \(0.5\).
• Calculate \(\frac{1}{2}[\cosh(2 \times 0.5) - \cosh(0)]\).

The result will depict the exact area under the curve from \(x=0\) to \(x=0.5\). Understanding this process is essential for grasping how areas and physical quantities can be derived from functions.

Differentiation Verification

Verification through differentiation is a process in calculus where you use differentiation to confirm the solution to an integral. This technique is essential for ensuring the accuracy of your integration results. In this example, after integrating \(\sinh 2x\), we obtained \(\frac{1}{2} \cosh 2x + C\). Therefore, to verify, we differentiate this expression:\[ \frac{d}{dx} \left( \frac{1}{2} \cosh 2x + C \right) \].
The constant \(C\) disappears, and by applying the chain rule, the derivative becomes:

• The derivative of \(\cosh 2x\) is \(2 \sinh 2x\).
• Multiply by \(\frac{1}{2}\), leading to \(\sinh 2x\).

The resulting expression matches the original function within the integral, \(\sinh 2x\), confirming the correctness of the integration. This step of differentiation is a powerful tool to check your work and ensure your solutions correctly model the mathematical relationship you're investigating. It is a pivotal part of learning and applying calculus effectively.