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Chapter 7: Problem 49

Evaluate the integrals. $$\int \frac{\operatorname{sech} \sqrt{t} \tanh \sqrt{t} d t}{\sqrt{t}}$$

Short Answer

Expert verified
The integral evaluates to \(-2 \operatorname{sech}(\sqrt{t}) + C\).

Step by step solution

01

Substitution

Let us begin by making the substitution \( u = \sqrt{t} \), which implies that \( t = u^2 \). The differential \( dt \) can also be transformed by differentiating \( t \) with respect to \( u \), giving \( dt = 2u \, du \).
02

Substitute and Simplify

Substitute \( u = \sqrt{t} \) and \( dt = 2u \, du \) into the integral. The expression becomes \( \int \frac{\operatorname{sech} u \tanh u \cdot 2u \, du}{u} \). Simplifying gives \( 2 \int \operatorname{sech} u \tanh u \, du \).
03

Integral of Hyperbolic Functions

Recall the derivative of \( \operatorname{sech} u \) is \( -\operatorname{sech} u \tanh u \). Hence the integral \( \int \operatorname{sech} u \tanh u \, du = -\operatorname{sech} u + C \), where \( C \) is the constant of integration.
04

Final Integration

Substituting back, we have \( 2(-\operatorname{sech} u) + C = -2 \operatorname{sech}(\sqrt{t}) + C \) as the evaluated integral.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Substitution Method
The Substitution Method is a powerful integration technique used to simplify complex integrals. It involves substituting a part of the integral with a new variable, which transforms the integral into a simpler form.
This method is particularly useful when the integral contains a composite function.

To perform substitution, follow these steps:
  • Identify a part of the integral to substitute with a new variable. In our example, we chose to let \( u = \sqrt{t} \). This simplifies the expression because the variable \( u \) streamlines the integration process.
  • Express the original variable in terms of the new variable, \( t = u^2 \) in this case. Substitute into the integral.
  • Calculate the differential of the substituted variable and replace the original differential with it. We found \( dt = 2u \, du \).
Substitution often helps in transforming the integral into a standard form that is easier to integrate.
Hyperbolic Functions
Hyperbolic functions are analogs of the trigonometric functions but for a hyperbola. The most common hyperbolic functions are \( \cosh(u) \) and \( \sinh(u) \). In our solution, we deal with \( \operatorname{sech}(u) \) and \( \tanh(u) \). These functions have unique derivatives, making them useful in solving integrals.

Understanding the properties of hyperbolic functions aids in integration:
  • Hyperbolic secant: \( \operatorname{sech}(u) = \frac{1}{\cosh(u)} \)
  • Hyperbolic tangent: \( \tanh(u) = \frac{\sinh(u)}{\cosh(u)} \)
  • Derivative of \( \operatorname{sech}(u) \) is \( -\operatorname{sech}(u) \tanh(u) \)
In our integral, knowing that the derivative of \( \operatorname{sech}(u) \) involves \( \tanh(u) \) simplifies the integration process since we can use this derivative to find the integral directly.
Definite and Indefinite Integrals
Integrals are fundamental in calculus and can be classified as definite or indefinite.
Indefinite integrals represent a family of functions and include a constant of integration, commonly denoted as \( C \). They are crucial for solving differential equations and finding antiderivatives.

In the given problem, we evaluated an indefinite integral, as shown by the presence of the constant \( C \) in the final solution.

Important points to remember:
  • Indefinite integrals don鈥檛 have upper or lower limits on the integral sign. For example, \( \int f(x) \,dx = F(x) + C \).
  • Definite integrals produce a numerical value, signifying the area under a curve, and have specific limits, which are not present in this problem.
Understanding these differences is vital when evaluating integrals, ensuring clarity whether you're determining a function family or calculating a specific area.

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