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Chapter 12: Problem 28

Prove: \(\int \operatorname{sech} u d u=2 \tan ^{-1} e^{u}+C\)

Short Answer

Expert verified
\( \int \operatorname{sech} u \, du = 2 \tan^{-1} e^u + C \)

Step by step solution

01

Recognize the integral to be solved

The integral to be solved is \( \int \operatorname{sech} u \, du \).
02

Rewrite the integrand

Recall that \( \operatorname{sech} u = \frac{2}{e^u + e^{-u}} \).
03

Use a substitution

Let \( v = e^u \). Then \( dv = e^u \, du \) or \( du = \frac{dv}{v} \). Substitute this in the integral: \( \int \operatorname{sech} u \, du = \int \frac{2}{v + \frac{1}{v}} \cdot \frac{dv}{v} \).
04

Simplify the integral

Simplify the expression: \( \int \frac{2}{v + \frac{1}{v}} \cdot \frac{dv}{v} = \int \frac{2}{\frac{v^2 + 1}{v}} \cdot \frac{dv}{v} = \int \frac{2v}{v^2 + 1} \cdot \frac{dv}{v} = 2 \int \frac{dv}{v^2 + 1} \).
05

Integrate

Recognize that \( \int \frac{dv}{v^2 + 1} = \tan^{-1} v \). Thus, \( 2 \int \frac{dv}{v^2 + 1} = 2 \tan^{-1} v \).
06

Substitute back

Substitute back \( v = e^u \) to get \( 2 \tan^{-1} e^u \).
07

Add the constant of integration

Thus, the integral is \( 2 \tan^{-1} e^u + C \).

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Key Concepts

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

hyperbolic functions
Hyperbolic functions are analogs of the trigonometric functions but for the hyperbola. They include \operatorname{sech}\, hyperbolic cosine \( \text{cosh} u \), hyperbolic sine \( \text{sinh} u \), and others. In this exercise, we worked with the hyperbolic secant, written as \operatorname{sech} u\. It is defined as:
\[ \operatorname{sech} u = \frac{2}{e^u + e^{-u}} \]
This similarity between hyperbolic and trigonometric functions often makes them appear in integral calculus just like their circular counterparts when solving integrals or differential equations.
integration by substitution
Integration by substitution is a technique used to transform a complex integral into a simpler one. It's like the reverse of the chain rule in differentiation. For the given problem, the substitution was made by setting \( v = e^u \). This helps to rewrite the integrand in a more manageable way.
  • The substitution choice simplifies the integral considerably.
  • After substituting, it鈥檚 important to also transform the differential \du\.
  • The new variable (\

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Most popular questions from this chapter

Find the derivative of the given function.\(F(x)=\tan ^{-1}\left(\sinh x^{2}\right)\)

Find the derivative of the given function.\(F(x)=\tanh ^{-1} 4 x\)

Prove: (a) \(\int \tanh u d u=\ln |\cosh u|+C ;\) (b) \(\int \operatorname{coth} u d u=\ln |\sinh u|+C\)

Prove the identities.(a) \(\tanh (-x)=-\tanh x ;\) (b) \(\operatorname{sech}(-x)=\operatorname{sech} x\)

Express the indefinite integral in terms of an inverse hyperbolic function and as a natural logarithm.\(\int \frac{d x}{\sqrt{x^{2}-4 x+1}}\)
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