Integration problem using substitution

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Homework Help Overview

The discussion revolves around an integration problem involving the substitution method, specifically using \( u = \sin 4x \) to evaluate the integral \( \int \cos^3 4x \, dx \). Participants explore various approaches to handle the integration and the implications of changing variables.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss different methods of substitution and integration, questioning whether to integrate with respect to \( u \) or \( x \). There are attempts to clarify the relationship between the variables and the limits of integration. Some participants express confusion about integrating one variable with respect to another.

Discussion Status

The discussion is ongoing, with participants providing guidance on ensuring all references to \( x \) are converted to \( u \). There is recognition of the need to express limits in terms of \( u \), and some participants have noted their confusion regarding the integration process. Multiple interpretations of the problem are being explored.

Contextual Notes

There are mentions of limits for the integral, with some participants correcting themselves on the values. The discussion also highlights the challenge of expressing trigonometric identities in terms of the substitution variable.

chwala
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Homework Statement



using ## u= sin 4x## find the exact value of ##∫ (cos^3 4x) dx##[/B]

Homework Equations

The Attempt at a Solution



## u= sin 4x## [/B]on integration ##u^2/2=-cos4x/4 ## , →##-2u^6={cos 4x}^3 ##...am i on the right track because now i end up with ##∫{{-2u^6}/{4.-2u^2}}du## or should i use

##du=4cos 4x dx ## to end up with ## 0.25 ∫ cos^24x du## which looks wrong to me
 
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chwala said:

## u= sin 4x##
on integration ##u^2/2=-cos4x/4 ## ,
No, you've integrated one side wrt u and the other wrt x.
 
haruspex said:
No, you've integrated one side wrt u and the other wrt x.
so should i use the second approach?
 
chwala said:
so should i use the second approach?
Yes, but you need to get all of the references to x turned into references to u.

You are asked for an exact value, but it is an indefinite integral. Remember that the limits need to be expressed in terms of u as well.
 
yes the limits are from 0 to π/24
 
is ## 0.25∫{cos^24x}du## correct?
 
chwala said:
the limits are from 0 to π/24
So what are the limits on u?
chwala said:
is ## 0.25∫{cos^24x}du## correct?
Yes.
 
haruspex said:
So what are the limits on u?

Yes.
limits on u are 0 to 30, now how do i proceed with the integration?
 
## cos^2 4x## = ##(cos 8x+1)##/2

should we substitute again? or are we going to have##0.25∫cos^24x d {sin4x} ##
 
  • #10
haruspex said:
So what are the limits on u?

Yes.
i am a bit confused we cannot integrate a variable say ##x## with respect to another variable say ##u##, i am stuck here
 
  • #11
chwala said:
limits on u are 0 to 30,
No.
chwala said:
how do i proceed with the integration?
You have the cos2 of some angle, and you need to express that in terms of u, the sine of the same angle. Does nothing click?
 
  • #12
haruspex said:
No.

You have the cos2 of some angle, and you need to express that in terms of u, the sine of the same angle. Does nothing click?
sorry limits are from 0 to 0.5 an oversight on my part...
 
  • #13
haruspex said:
No.

You have the cos2 of some angle, and you need to express that in terms of u, the sine of the same angle. Does nothing click?
i now get it lol
## 0.25∫{1-u^2}du ## from u=0 to u=0.5 thanks mate solution is ## 0.115##
 
  • #14
Why don't you try splitting ##cos^34x## into ##cos4x## and another term containing the term used for ##u## substitution?
 
  • #15
Eclair_de_XII said:
Why don't you try splitting ##cos^34x## into ##cos4x## and another term containing the term used for ##u## substitution?
i have seen the obstacle with that move...
 
  • #16
chwala said:
i have seen the obstacle with that move...
i have seen it, check post 13
 
  • #17
Eclair_de_XII said:
Why don't you try splitting ##cos^34x## into ##cos4x## and another term containing the term used for ##u## substitution?
chwala said:
i have seen the obstacle with that move...
@chwala
Actually that is the move you finally made to solve. Check the time of @Eclair_de_XII 's post and your posts.
 

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