Problem 47 - Calculus of a Single Variable

Question

In Exercises 43–54, find the indefinite integral. $$ \int \cosh ^{2}(x-1) \sinh (x-1) d x $$

Step-by-Step Solution

Answer

The indefinite integral of \(\int \cosh ^{2}(x-1) \sinh (x-1) d x\) is \(\cosh(x-1) + \frac{\cosh^3(x-1)}{3} - \cosh(x-1)\).

Step 1: Use trigonometric identity

Start by expressing the function \(\cosh^2(x-1)\) in terms of \(\sinh(x-1)\) using the identity \(\cosh^2(x) - \sinh^2(x) = 1\). The integral then becomes:\[ \int (1+\sinh^2(x-1))\sinh(x-1) d x \]

Step 2: Distribute and separate

Next, distribute the \(\sinh(x-1)\) term and separate the integral into two parts:\[ \int \sinh(x-1)dx + \int \sinh^3(x-1)dx \]

Step 3: Simplify first integral

The first integral on the right is simply the integral of \(\sinh(x-1)\), which integrates to \(\cosh(x-1)\). We then have:\[ \cosh(x-1) + \int \sinh^3(x-1)dx \]

Step 4: Break down the second integral

To solve the second integral, break \(\sinh^3(x-1)\) into \(\sinh(x-1)\) and \(\sinh^2(x-1)\), then substitute \(\sinh^2(x-1)\) with \(\cosh^2(x-1) -1\). The second integral then becomes:\[ \int [\sinh(x-1) (\cosh^2(x-1)-1)] dx \]

Step 5: Evaluate the second integral

This integral can now be evaluated using substitution method. If we let \(u=\cosh(x-1)\), then \(du=\sinh(x-1) dx\). The integral then becomes\[ \int [u^2 - 1] du \] which integrates to:\[ \frac{u^3}{3} - u \] Substitute back \(u\) with \(\cosh(x-1)\) and this gives \(\frac{\cosh^3(x-1)}{3} - \cosh(x-1)\).

Step 6: Add the solutions

Finally, add the solutions from the two integrals we evaluated:\[ \cosh(x-1) + \frac{\cosh^3(x-1)}{3} - \cosh(x-1) \] Simplify and this gives the final primitive function.

Key Concepts

Trigonometric Identity

When we work with integrals involving trigonometric or hyperbolic functions, it's often useful to apply trigonometric identities that can simplify the integral or make it more approachable. A common identity used in the context of hyperbolic functions is \[\begin{equation} \cosh^2(x) - \sinh^2(x) = 1 \end{equation}\]It's analogous to the Pythagorean identity \( \sin^2(x) + \cos^2(x) = 1 \) used with circular trigonometric functions. These identities are extremely helpful in breaking down complex expressions into forms that are easier to integrate. In our example, the identity transforms the original integral of \(\cosh^2(x-1)\sinh(x-1)\) into a sum of two simpler integrals, allowing for a more straightforward integration process.

Hyperbolic Functions

Just like their trigonometric counterparts, hyperbolic functions, including \(\sinh(x)\), \(\cosh(x)\), and \(\tanh(x)\), have properties and identities that can be exploited to solve integrals. These functions are defined using exponential functions:

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

Their derivatives are interrelated too, with \(\frac{d}{dx}\cosh(x) = \sinh(x)\) and \(\frac{d}{dx}\sinh(x) = \cosh(x)\), making them particularly amenable to integration and differentiation in calculus.

Integration Techniques

In calculus, there are several techniques for finding integrals, and knowing which one to apply is key to solving problems efficiently. Some common techniques include the power rule, integration by parts, trigonometric substitution, and partial fraction decomposition. In complex cases, we may have to combine these methods or perform additional algebraic manipulation to make an integral solvable. For integrals of hyperbolic functions, we often rely on their unique identities and differentiation properties to find their antiderivatives.

Substitution Method

The substitution method, also known as u-substitution, is a technique where you choose a part of the integral to represent with a new variable, \(u\), which simplifies the integral into a form that is easier to evaluate. The choice of \(u\) is often guided by looking for a function within the integral whose derivative is also present. In our example, the choice of \(u = \cosh(x-1)\) simplifies the integral of \(\sinh^3(x-1)\) dramatically because the derivative of \(\cosh(x-1)\), which is \(\sinh(x-1)\), appears outside of the cubed term in the integral. The process involves differentiating \(u\) to get \(du\), and then replacing parts of the original integral with \(u\) and \(du\) to integrate with respect to \(u\) instead of the original variable.