CalculusLesson 5 of 5 in the course Techniques of integration

Reduction Formulae for Cosine Powers and Polynomial–Exponential Integrals

5 flashcards · 5 quiz questions · 5 match pairs · 5 fill-in sentences · English

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Summary

A reduction formula turns a difficult family of integrals into a recurrence: an integral with a higher integer parameter is expressed using one with a lower parameter. Repeating the relation eventually reaches a simple base integral.

Suppose a family of integrals depends on an integer nn. We write it as

A reduction formula expresses InI_n using a lower-index integral, such as In−1I_{n-1} or In−2I_{n-2}, together with any necessary algebraic terms. The lower index represents a simpler member of the same family.

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What is a reduction formula?
A relation that expresses an integral with a higher parameter using lower-parameter integrals and possibly known terms.
What does a base case do in a reduction process?
It provides a directly evaluable integral where repeated reduction can stop.
Cosine-power reduction formula
For In=∫cos⁡nx dxI_n=\int \cos^n x\,dx, In=sin⁡xcos⁡n−1xn+n−1nIn−2I_n=\frac{\sin x\cos^{n-1}x}{n}+\frac{n-1}{n}I_{n-2}.
Polynomial–exponential reduction formula
For In=∫xneax dxI_n=\int x^n e^{ax}\,dx with a≠0a\ne 0, In=xneaxa−naIn−1I_n=\frac{x^n e^{ax}}{a}-\frac{n}{a}I_{n-1}.
Back-substitution
Using lower-index results to build the value of the original integral.

Quiz questions

  1. For In=∫cos⁡nx dxI_n=\int \cos^n x\,dx, which recurrence reduces the exponent by two?Answer: In=sin⁡xcos⁡n−1xn+n−1nIn−2I_n=\frac{\sin x\cos^{n-1}x}{n}+\frac{n-1}{n}I_{n-2}
  2. For In=∫xneax dxI_n=\int x^n e^{ax}\,dx, where a≠0a\ne 0, which recurrence follows from integration by parts?Answer: In=xneaxa−naIn−1I_n=\frac{x^n e^{ax}}{a}-\frac{n}{a}I_{n-1}
  3. For In=∫xneax dxI_n=\int x^n e^{ax}\,dx, what should be done when a=0a=0?Answer: Integrate the resulting power xnx^n directly.
  4. Repeatedly applying the cosine-power recurrence to I5I_5 reaches which base integral?Answer: I1=∫cos⁡x dxI_1=\int \cos x\,dx
  5. To derive a recurrence for In=∫xneax dxI_n=\int x^n e^{ax}\,dx by integration by parts, which choice is appropriate?Answer: u=xnu=x^n and dv=eax dxdv=e^{ax}\,dx

Based on the Wikipedia article Integration by reduction formulae, licensed CC BY-SA 4.0. Adapted into a summary, flashcards and quizzes.

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