How to Integrate Cotangent and Cosecant with Odd Powers?

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SUMMARY

The integration of cotangent and cosecant functions with odd powers can be approached by first rewriting cot^a(x) and csc^b(x) in terms of sine and cosine. Specifically, cot^a(x) is expressed as cos^a(x) / sin^a(x) and csc^b(x) as 1/sin^b(x). The integration strategy involves removing one factor each of csc(x) and cot(x), leading to the integral ∫ cot^{a-1}(x)csc^{b-1}(x) csc(x)cot(x) dx. Utilizing the identity cot^2(x) + 1 = csc^2(x) allows for substitution, ultimately simplifying the integral into a manageable form.

PREREQUISITES
  • Understanding of trigonometric identities, specifically cotangent and cosecant functions.
  • Familiarity with integration techniques, including substitution methods.
  • Knowledge of how to manipulate powers of trigonometric functions.
  • Ability to apply u-substitution in calculus.
NEXT STEPS
  • Study the application of trigonometric identities in integration, focusing on cotangent and cosecant.
  • Learn advanced integration techniques, particularly those involving odd powers of trigonometric functions.
  • Explore the use of u-substitution in calculus with a focus on trigonometric integrals.
  • Practice integrating functions involving cotangent and cosecant using various methods.
USEFUL FOR

Students of calculus, mathematicians, and educators looking to deepen their understanding of trigonometric integration techniques, particularly those involving odd powers of cotangent and cosecant.

adelaide87
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Describe the method you would use to integrate

cot^a(x) csc^b(x) dx

If a and b are odd?

An explanation of the strategy would be a huge help!
 
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I would definitely try to put everything into terms of sin and cos.

cot^a(x) = Cos^a(x) / sin^a(x)

csc^b(x) = 1/sin^b(x)

this is where i would start. from there, you can sum the powers of the sin's in the denominators and try u substitution or something.
 
adelaide87 said:
Describe the method you would use to integrate

cot^a(x) csc^b(x) dx

If a and b are odd?

An explanation of the strategy would be a huge help!
Take out one factor each of csc x and cot x, leaving you with
\int cot^{a-1}(x)csc^{b - 1}(x)~csc(x)cot(x)dx

Use the identity cot2(x) + 1 = csc2(x) (or equivalently, cot2(x) = csc2(x) - 1) to replace the cota - 1 factor.

At that point you'll have a sum of terms that involve various powers of csc2(x) and you can use an ordinary substitution, with u = csc(x).
 

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