Problem 39 - Essential Calculus

Question

Find the integral. \(\int \cosh ^{2}(x-1) \sinh (x-1) d x\)

Step-by-Step Solution

Answer

The integral of \( \cosh^2(x-1) \sinh(x-1) \) is \( \cosh^2(x-1)^2 + c \)

Step 1: Rewrite the Integral

First, recognize that \(\sinh(x-1)\) is the derivative of \(\cosh(x-1)\). So we can rewrite the integral as \(\int u du\) where \(u = \cosh^2(x-1)\). Then \(\frac{du}{dx} = 2 \cosh(x-1) \sinh(x-1)\). Notice that we can now cancel out the \(\sinh(x-1)\) terms.

Step 2: Change of Variables

To perform a change of variables, it might be easier to bring the constant factor of 2 outside of the integral. Now we have \(2 \int \cosh(x-1) du\). Referring to our substitution, this just turns into \(2u du\).

Step 3: Compute the Integral

Evaluate the integral, leaving us with: \( u^2+c \). Substitute back for \( u \) giving us the final answer \(\cosh^2(x-1)^2 + c \).

Key Concepts

Hyperbolic Functions

Hyperbolic functions, like the ones encountered in the exercise, are mathematical functions that share similar properties to trigonometric functions. However, instead of relating to circles, hyperbolic functions relate to hyperbolas. The two primary hyperbolic functions are the hyperbolic sine, written as \( \sinh(x) \), and hyperbolic cosine, written as \( \cosh(x) \).

• The function \( \sinh(x) \) is defined as \( \sinh(x) = \frac{e^x - e^{-x}}{2} \).
• Conversely, \( \cosh(x) \) is defined as \( \cosh(x) = \frac{e^x + e^{-x}}{2} \).

An important property of these functions is their derivatives:

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

This property is crucial in integration, as it allows us to identify when we might use techniques like substitution to simplify a problem.

Substitution Method

The substitution method is a common technique in integral calculus that simplifies an integral by changing variables. Essentially, we replace a part of the integrand with a new variable. This technique is particularly helpful when the integrand involves a function and its derivative.
In the provided exercise, the substitution method was used by recognizing \( \sinh(x-1) \) as the derivative of \( \cosh(x-1) \). Here’s a quick rundown of how substitution was applied:

• Recognize the derivative: We noticed that \( \sinh(x-1) \) is the derivative of \( \cosh(x-1) \).
• Choose a substitution variable: Let \( u = \cosh^2(x-1) \).
• Compute \( \frac{du}{dx} \): This gave \( 2 \cosh(x-1)\sinh(x-1) \), which allowed us to perform the substitution and simplify the integral.
• Integrate with respect to \( u \): Once the substitution is done, the integral becomes simpler and easier to solve.

After integrating in terms of the new variable, don't forget to substitute back the original function to get the final answer in terms of \( x \). This completes the process.

Integral Calculus

Integral calculus is a fundamental branch of calculus that deals with integration, which is essentially the reverse operation of differentiation. The goal is to find a function (the antiderivative) whose derivative is the given function.

• An integral can give us the area under a curve within certain bounds (definite integral), or an antiderivative function (indefinite integral).
• In the exercise provided, finding the indefinite integral was our goal, meaning we sought the antiderivative of the given expression.

Key aspects to keep in mind include:

• The process involves reversing differentiation, so tools like substitution (as described previously) become handy.
• Often, integration involves recognizing forms that match standard integral rules or simplify through algebraic manipulation.

By understanding integral calculus, including techniques like substitution and recognition of function patterns, you can tackle a wide range of problems involving integration. It's through such methods that complex-looking integrals, such as those involving hyperbolic functions, become manageable.