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Chapter 7: Problem 22

Evaluate the integrals. $$\int e^{\csc (\pi+t)} \csc (\pi+t) \cot (\pi+t) d t$$

Short Answer

Expert verified
The integral evaluates to \(-e^{\csc(\pi+t)} + C\).

Step by step solution

01

Identify Substitution Component

Notice that the integral involves an exponential function with a trigonometric argument. Typically, for integrals of the form \( e^u u' \, dt \), we use substitution. Here, we have \( u = \csc(\pi + t) \), whose derivative is \( -\csc(\pi + t) \cot(\pi + t) \). This matches part of the integral.
02

Perform Substitution

Set \( u = \csc(\pi + t) \) and find \( du = -\csc(\pi + t) \cot(\pi + t) \, dt \). Notice that \( du = -d(t) \textb{ as } \csc(\pi + t) \cot(\pi + t) \, dt = -du \). To adapt the integral, multiply through by \(-1\), transforming our integral to \(-\int e^u \, du\).
03

Integrate the Expression

Now, integrate \(-\int e^u \, du\), which is a standard result: \(-e^u + C\). This is because the integral of \(e^u\) is \(e^u\), with the \(-1\) multiplicative constant included from the substitution adjustment.
04

Back-Substitute to Original Variable

Replace \(u\) with the original substitution: \(u = \csc(\pi + t)\). Hence, the integral becomes \(-e^{\csc(\pi + t)} + C\). This back-substitution returns the solution to the variable in the original integral.

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Key Concepts

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

Trigonometric Substitution
Trigonometric substitution is a technique used in calculus to simplify integrals by transforming complex trigonometric expressions into simpler algebraic forms. In this approach, we replace a trigonometric function with a variable, easing the integration process. In the given problem, we notice the presence of a trigonometric component, specifically, the cosecant
  • First, identify a trigonometric function in the problem that complicates direct integration. Here, it's \( \csc(\pi + t) \).
  • Then, set this function equal to a new variable, typically denoted as \(" u "\). For this exercise, \{ u = \csc(\pi + t) \}.
  • Find the derivative of this new variable in terms of the original variable. This means calculating \{ du = -\csc(\pi + t) \cot(\pi + t) \, dt \}.
By using trigonometric substitution, we essentially transform the integral into a more manageable form, paving the way for straightforward integration techniques.
Exponential Functions
Exponential functions appear frequently in calculus, representing rapidly increasing or decreasing curves. They have the form
  • The exponent can be any expression, which in this problem is \( e^{\csc(\pi+t)} \).
When paired with trigonometric functions, they often require substitution for easier integration.
  • Recognize exponential expressions of the form \( e^u \). These are immediately integrable to \( e^u \), up to a constant adjustment.
  • After substituting \( u = \csc(\pi + t) \), the integral simplifies to \(- \int e^u \, du\).
This highlights the simplicity of integrating standard exponential forms once the substitution transforms the integral. Ultimately, these functions simplify to elegant mathematical expressions through strategic replacements.
Indefinite Integral
An indefinite integral is essentially the opposite of differentiation, representing a family of functions that alleviate the derivative back to an original function plus a constant. The symbol
  • It encompasses all possible functions that could result in the original function's derivative, noted by the constant \( C \).
  • For an expression like \( \int e^{\csc(\pi+t)} \csc(\pi+t) \cot(\pi+t) \, dt \), it involves reversing the differentiation process in a structured manner.
  • After acknowledging the original substitution, integrate the transformed expression, where \( \int e^u \, du \) resolves to \( e^u + C \), and multiply by any constants derived from previous steps, notably the \(-1\) from \(-du\).
Returning to the original variables after integrating, the constant \( C \) ensures all potential solutions of the integral are captured, reaffirming the broader applicability of the indefinite integral.

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Most popular questions from this chapter

Evaluate the integrals. $$\int_{-\ln 2}^{0} \cosh ^{2}\left(\frac{x}{2}\right) d x$$

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