Problem 36 - Calculus Early Transcendentals

Question

Determine each indefinite integral. \(\int \sinh ^{2} x d x(\text { Hint: Use an identity. })\)

Step-by-Step Solution

Answer

Question: Find the indefinite integral of the hyperbolic sine function squared, \(\int \sinh^2x\,dx\). Answer: \(\int \sinh^2x\,dx = \frac{1}{4}\sinh(2x) - \frac{1}{2}x + C\), where \(C\) is the constant of integration.

Step 1: Applying the identity

We have the identity: \(\cosh(2x) = \cosh^2x - \sinh^2x\). With some rearrangement, we can find the expression for \(\sinh^2x\): \(\sinh^2x = \cosh(2x) - \cosh^2x\) Now we can rewrite the given integral as: \(\int \sinh^2x\,dx = \int (\cosh(2x) - \cosh^2x)\,dx\)

Step 2: Separating the integral

We can separate the integral into two parts: \(\int (\cosh(2x) - \cosh^2x)\,dx = \int \cosh(2x)\,dx - \int \cosh^2x\,dx\)

Step 3: Integrate the first part

Now we evaluate the integral of \(\cosh(2x)\). \(\int \cosh(2x)\,dx = \frac{1}{2} \sinh(2x) + C_1\)

Step 4: Integrate the second part

To evaluate the integral of \(\cosh^2x\), we need another identity: \(\cosh(2x) = 2 \cosh^2x - 1\). Rearranging for \(\cosh^2x\), we get: \(\cosh^2x = \frac{1}{2}(\cosh(2x) + 1)\) Now, we can rewrite the integral as: \(\int \cosh^2x\,dx = \int \frac{1}{2}(\cosh(2x) + 1)\,dx\) We can now evaluate this integral easily: \(\int \frac{1}{2}(\cosh(2x) + 1)\,dx = \frac{1}{4}\sinh(2x) + \frac{1}{2}x + C_2\)

Step 5: Combine the results

Now we have the indefinite integrals of both parts. We can combine them to find the indefinite integral of the original expression: \(\int \sinh^2x\,dx = \int (\cosh(2x) - \cosh^2x)\,dx\) \(= \int \cosh(2x)\,dx - \int \cosh^2x\,dx\) \(= \frac{1}{2}\sinh(2x) + C_1 - \left(\frac{1}{4}\sinh(2x) + \frac{1}{2}x + C_2\right)\) \(= \frac{1}{4}\sinh(2x) - \frac{1}{2}x + C\) Here, \(C = C_1 - C_2\) is the constant of integration.

Key Concepts

Understanding Hyperbolic Functions

Hyperbolic functions are analogous to trigonometric functions but are based on hyperbolas rather than circles. The two primary hyperbolic functions are \(\sinh(x)\) and \(\cosh(x)\). These functions are defined as follows:

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

Because hyperbolic functions are closely related to exponential functions, they exhibit properties similar to trigonometric identities. These properties are useful in various mathematical applications, including solving differential equations and performing integrals.

Notice that these functions have relationships similar to trigonometric functions. For example, the identity \(\cosh^2x - \sinh^2x = 1\) resembles the Pythagorean identity \(\cos^2x + \sin^2x = 1\). These types of identities can be key in simplifying expressions when integrating functions involving hyperbolic sine and cosine.

Exploring Integration Techniques

Integral calculus is a powerful tool used to find areas under curves, among other applications. To solve integrals like the indefinite integral of \( \sinh^2x \), we often use integration techniques that involve algebraic manipulation and trigonometric or hyperbolic identities.

One common technique is substitution, where you replace part of the integral with another variable to simplify it. However, in this exercise, the use of identities plays a crucial role. By expressing \( \sinh^2x \) in terms of \( \cosh(2x) \) and \( \cosh^2x \), the integral becomes more manageable. This is done through the identity:

\[ \cosh(2x) = \cosh^2x - \sinh^2x \]
You rearrange this identity to express \( \sinh^2x \) in terms of \( \cosh(2x) \) and \( \cosh^2x \) and then segregate the integral into simpler parts.

This step allows us to support the integration process with known antiderivatives like \( \int \cosh(2x)dx \) and turns the problem into a relatively straightforward computation. Understanding how these techniques converge is vital for solving more complex integral problems.

Utilizing Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for all values of the involved variables. Similar concepts apply to hyperbolic functions, as seen in this exercise where we use identities to simplify expressions before integration.

For integrals involving \( \sinh(x) \) or \( \cosh(x) \), using identities can make a seemingly complex integral into a simpler form. Consider, for example, the identity \( \cosh(2x) = 2\cosh^2x - 1 \), used to express \( \cosh^2x \) in terms of \( \cosh(2x) \) and 1, helping to dismantle the integral into a combination of easier-to-manage pieces.

Similarly, the identity \( \cosh^2x - \sinh^2x = 1 \) mirrors the trigonometric Pythagorean identity and strengthens our ability to transform the integral to a form where standard integration rules can be applied.

Learning these identities and understanding when and how to employ them is a critical skill for students tackling trigonometric or hyperbolic integrals and offers significant problem-solving efficiency.