Evaluating $\int_{-1}^{1}\sin^7\left({x}\right) \,dx$ Using $u$-Substitution

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Discussion Overview

The discussion revolves around evaluating the integral $\int_{-1}^{1}\sin^7\left({x}\right) \,dx$ using $u$-substitution and exploring properties of odd functions in definite integrals. The scope includes mathematical reasoning and conceptual clarification regarding symmetry in integrals.

Discussion Character

  • Mathematical reasoning, Conceptual clarification, Debate/contested

Main Points Raised

  • One participant proposes using the substitution $u=\cos\left({x}\right)$ and expresses a desire to continue the evaluation.
  • Another participant points out that the integrand is an odd function and suggests considering a shortcut due to the symmetry of the limits.
  • A participant interprets the suggestion about symmetry, presuming that the areas will cancel out, leading to a potential value of zero for the integral.
  • A later reply confirms the application of the odd-function rule, stating that the definite integral evaluates to zero.

Areas of Agreement / Disagreement

Participants generally agree on the application of the odd-function property, leading to the conclusion that the integral evaluates to zero. However, there is a lack of consensus on the necessity of the substitution method presented initially.

Contextual Notes

The discussion does not resolve the implications of the substitution method versus the odd-function property, leaving the effectiveness of both approaches open to interpretation.

karush
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$\int_{-1}^{1}\sin^7\left({x}\right) \,dx$
$u=\cos\left({x}\right)$
$du=-\sin\left({x}\right)$
So
$-\int_{-1}^{1} \left(1-{u}^{2}\right)^3\,du$
So continue?
 
Last edited:
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Your integrand is odd, and the limits symmetric about the origin, so can you think of a shortcut you can use?
 
I presume you mean the areas will cancel out and the answer will be zero for interval given?
 
karush said:
I presume you mean the areas will cancel out and the answer will be zero for interval given?

Yes, the odd-function rule can be directly applied here to state that the given definite integral has a value of 0. :D
 

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