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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 7, Problem 7.PE.47

Evaluate the integrals in Exercises 31–78.
47. ∫(1/r)csc²(1+ln(r))dr

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1
Identify the integral to solve: \(\int \frac{1}{r} \csc^{2}(1 + \ln(r)) \, dr\).
Recognize that the integrand involves a composite function \(1 + \ln(r)\) inside the \(\csc^{2}\) function, and the factor \(\frac{1}{r}\) outside suggests a substitution related to \(\ln(r)\).
Let \(u = 1 + \ln(r)\). Then, compute the differential \(du\): since \(\frac{d}{dr} \ln(r) = \frac{1}{r}\), it follows that \(du = \frac{1}{r} dr\).
Rewrite the integral in terms of \(u\): substitute \(\frac{1}{r} dr\) with \(du\), so the integral becomes \(\int \csc^{2}(u) \, du\).
Recall the antiderivative of \(\csc^{2}(u)\) is \(-\cot(u) + C\). After integrating, substitute back \(u = 1 + \ln(r)\) to express the answer in terms of \(r\).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Integration by Substitution

Integration by substitution is a method used to simplify integrals by changing variables. It involves identifying a part of the integrand as a new variable, differentiating it, and rewriting the integral in terms of this variable. This technique is especially useful when the integral contains a composite function, such as csc²(1 + ln(r)).
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Properties of Logarithmic Functions

Logarithmic functions, like ln(r), have specific differentiation and integration properties. Understanding that the derivative of ln(r) is 1/r helps in recognizing substitution candidates. This knowledge is crucial when the integrand involves expressions like 1/r combined with functions of ln(r).
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Integration of Trigonometric Functions

Integrals involving trigonometric functions such as csc²(x) rely on knowing their antiderivatives. For example, the integral of csc²(x) dx is -cot(x) + C. Recognizing these standard integrals allows for straightforward evaluation once the integral is properly transformed.
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