Integrate cos^3xdx - First Clue: Substitute (1-(Sinx)^2)cosxdx

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Discussion Overview

The discussion revolves around techniques for integrating the function \(\int \cos^3 x \, dx\) and related integrals of the form \(\int \cos^n x \, dx\) for odd integers \(n\) from 3 to 13. Participants explore substitution methods and seek validation for their integration steps.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant suggests starting with the substitution \((1 - \sin^2 x) \cos x \, dx\) for \(\int \cos^3 x \, dx\).
  • Another participant questions the applicability of learned techniques to the integral \(\int (1 - \sin^2 x) \cos x \, dx\).
  • A participant proposes letting \(u = \sin x\) to transform the integral into \(\int (1 - u^2)^2 \, du\) and provides a detailed breakdown of the integration process.
  • Concerns are raised about potential errors in the integration steps, particularly regarding the substitution and simplification.
  • Some participants express differing views on the correctness of the integration steps, with one asserting a mistake in the substitution process and suggesting a general reduction formula approach.
  • Another participant agrees with the correctness of the integration but notes the importance of substituting back to the original variable.

Areas of Agreement / Disagreement

Participants express differing opinions on the validity of the integration steps presented, with some affirming correctness and others pointing out potential errors. The discussion remains unresolved regarding the accuracy of the proposed methods.

Contextual Notes

Participants discuss specific substitution techniques and simplifications, but there are unresolved issues regarding the correctness of these steps and the general approach to integrating powers of cosine.

lastlaugh
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Any suggestions on how to integrate
\int cos^3xdx
My first clue would be to substitute
(1-(Sinx)^2)cosxdx but any clue to go after this?
 
Last edited:
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Well, do any of the techniques you've learned apply to

\int (1 - \sin^2 x) \cos x \, dx

?
 
Put sinx=t => cosx dx=dt

so u have int(1-t2)dt=t-t3/3
 
thanks for the help. This trig integration stuff is killer

OK the reason I asked was because I have a lab assignment where I have to calculate the powers of \int cos^n[x]dx where n is odd numbers from 3-13.
So can someone please look over my work for \int cos^5[x]dx and make sure it looks valid:
First I substituted (1-sin^2[x])^2 for cos^4[x] which gives me:
\int (1-sin^2[x])^2 cos[x]dx
Then I let u=sinx meaning du=cosxdx
Giving me:
\int (1-u^2)^2du
Which I simplfied to give me:
\int u^4 du - \int 2u^2 du + \int du
Which gives me
(u^5)/5-(2u^3)/3 + u

Anyone see any glaring errors?
Thank you in advance.
 
Last edited:
Originally posted by lastlaugh
thanks for the help. This trig integration stuff is killer

OK the reason I asked was because I have a lab assignment where I have to calculate the powers of \int cos^n[x]dx where n is odd numbers from 3-13.
So can someone please look over my work for \int cos^5[x]dx and make sure it looks valid:
First I substituted (1-sin^2[x])^2 for cos^4[x] which gives me:
\int (1-sin^2[x])^2 cos[x]dx
Then I let u=sinx meaning du=cosxdx
Giving me:
\int (1-u^2)^2du
Which I simplfied to give me:
\int u^4 du - \int 2u^2 du + \int du
Which gives me
(u^5)/5-(2u^3)/3 + u

Anyone see any glaring errors?
Thank you in advance.



WRONGGGGGGGGGGGGGGGGGGGG

is (1-u^2) du.
 
Looks right to me. (except you should substitute back what u is)
 
Originally posted by PrudensOptimus
WRONGGGGGGGGGGGGGGGGGGGG

is (1-u^2) du.

How it is wrong

Moreover General way will be to form a reduction formula
 
Last edited:
Originally posted by Hurkyl
(except you should substitute back what u is)
Yeah I didn't feel like typing out any more latex:smile:
 

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