Problem 36 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

Question: Find the indefinite integral of \(\sinh^2 x\), given the hint that an identity can be used to simplify the integrand before integration. Answer: \(\int \sinh^2 x \, dx = \frac{1}{2}(\cosh x\sinh x) - x + C\)

Step 1: Recall the identity of hyperbolic functions

Recall that one of the identities for hyperbolic functions is: \(\cosh^2 x - \sinh^2 x = 1\) We can rearrange the formula to isolate \(\sinh^2 x\): \(\sinh^2 x = \cosh^2 x - 1\)

Step 2: Substitute the integrand with the identity

Replace the integrand with the identity derived in Step 1: \(\int \sinh^2 x \, dx = \int (\cosh^2 x - 1) \, dx\)

Step 3: Integrate each term separately

We can break up the integral and integrate each term separately: \(\int (\cosh^2 x - 1) \, dx = \int \cosh^2 x \, dx - \int 1 \, dx\)

Step 4: Integrate the first term

To integrate the first term, \(\int \cosh^2 x \, dx\), recall that the derivative of \(\sinh x\) is \(\cosh x\). Thus, we can use integration by parts, with \(u = \cosh x\) and \(dv = \cosh x \, dx\): $$ \int \cosh^2 x \, dx = u \int dv - \int \big(\frac{du}{dx} \int dv\big)dx = \cosh x\sinh x - \int (\sinh x)(\sinh x \, dx) $$ Notice that \(\int (\sinh x)(\sinh x \, dx) = \int \sinh^2 x \, dx\), which we can call \(I\): \(I = \cosh x\sinh x - I\) Now, solve for \(I\): \(2I = \cosh x\sinh x \Rightarrow I = \frac{1}{2}(\cosh x\sinh x)\)

Step 5: Integrate the second term

The integration of the second term is straightforward: \(\int 1 \, dx = x\)

Step 6: Combine the results

Now, combine the results from Steps 4 and 5: \(\int \sinh^2 x \, dx = \int (\cosh^2 x - 1) \, dx = \int \cosh^2 x \, dx - \int 1 \, dx = \frac{1}{2}(\cosh x\sinh x) - x + C\) Where \(C\) is the constant of integration. So the final answer is: \(\int \sinh^2 x \, dx = \frac{1}{2}(\cosh x\sinh x) - x + C\)

Key Concepts

Hyperbolic Functions

Hyperbolic functions, similar to trigonometric functions, are relationships that arise in hyperbolic geometry. They include functions such as sinh (hyperbolic sine) and cosh (hyperbolic cosine). These functions are defined using exponential functions:

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

One important identity involving hyperbolic functions is \( \cosh^2 x - \sinh^2 x = 1 \). This identity is similar to the Pythagorean identity used in trigonometry. It can be rearranged to express \( \sinh^2 x \) in terms of \( \cosh^2 x \), which becomes \( \sinh^2 x = \cosh^2 x - 1 \). This rearrangement is very useful in solving integrals involving hyperbolic functions.

Integration by Parts

Integration by parts is a useful technique for finding integrals, based on the product rule for differentiation. When faced with a product of functions, integration by parts can help break down the integral into more manageable parts. The formula is given as:\[ \int u \, dv = uv - \int v \, du \]

• Choose \( u \) and \( dv \) wisely. Let \( u \) be a function that becomes simpler when differentiated and \( dv \) a function whose integral is easily determined.
• The derivative \( du \) is found from \( u \), and \( v \) is the integral of \( dv \).

In this particular exercise, to integrate \( \cosh^2 x \), we marked \( u = \cosh x \) and \( dv = \cosh x \, dx \). Differentiating and integrating these respectively gives us the nice simplification to compute the integral effectively.

Integral Identities

Integral identities are essential in simplifying expressions within integrals. They provide a way to transform a complex integrand into a form that is easier to integrate. The hyperbolic identity \( \sinh^2 x = \cosh^2 x - 1 \) is an example of using an identity to simplify \( \int \sinh^2 x \, dx \). By substituting \( \sinh^2 x \) with \( \cosh^2 x - 1 \), the problem turns from one that seems complex into simpler integrals.

• Identity substitutions can greatly reduce the difficulty by transforming the original integral.
• Using identities can split the integral into parts that are more directly integrable.

This approach of applying identities allows mathematicians and students to solve problems that would otherwise require more complex methods.