Question on basic trig substitution with x = sin theta

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Discussion Overview

The discussion revolves around the use of trigonometric substitution in integrals, specifically the substitution of \( x = \sin(\theta) \) versus \( x = \cos(\theta) \). Participants explore the implications of these substitutions and their connection to the unit circle, as well as the general principles of variable substitution in calculus.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions the necessity of using \( x = \sin(\theta) \) instead of \( x = \cos(\theta) \), suggesting that the latter is more appropriate based on its representation on the unit circle.
  • Another participant argues that the choice of substitution is arbitrary and can be any invertible smooth function, emphasizing that it simplifies the integral rather than being tied to the unit circle's coordinates.
  • A participant references a lecture that suggests trigonometric substitutions are inherently linked to the unit circle, raising concerns about the validity of using either sine or cosine.
  • Some participants clarify that both substitutions rely on the identity \( \cos^2(\theta) + \sin^2(\theta) = 1 \), allowing for flexibility in choosing either function.
  • There is a discussion about the mechanics of variable substitution in integrals, with one participant providing a formal explanation of how the substitution process works mathematically.

Areas of Agreement / Disagreement

Participants express differing views on the appropriateness of using \( x = \sin(\theta) \) versus \( x = \cos(\theta) \) for trigonometric substitution. While some agree that either substitution is valid, others maintain that the choice is influenced by the context of the unit circle. The discussion remains unresolved regarding the implications of these choices.

Contextual Notes

Participants highlight that the choice of substitution may depend on the specific integral being evaluated and the resulting simplifications. There is also mention of the need for a one-to-one function in the substitution process, which may not have been fully explored in the discussion.

RoboNerd
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Say I have the integral of [ 1 / ( sqrt( 1 - x^2) ] * dx . Now I was told by many people in videos that I substitute x = sin theta, and this has me confused.

Wouldn't I need to substitute x = cos theta instead? as x = cos theta on the unit circle instead of sin theta?

Thanks in advance for your input! I am looking forward to hearing from you.
 
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It really does not matter. The substitution in itself has nothing to do with x being the x coordinate of the unit circle, it is just the name of a new integration variable. You can call it whatever you want, x, y, t, or ##\xi##, it is all the same.
 
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RoboNerd said:
Say I have the integral of [ 1 / ( sqrt( 1 - x^2) ] * dx . Now I was told by many people in videos that I substitute x = sin theta, and this has me confused.

Wouldn't I need to substitute x = cos theta instead? as x = cos theta on the unit circle instead of sin theta?

Thanks in advance for your input! I am looking forward to hearing from you.

Orodruin's reply is a very good one. The only thing you use is the identity cos^2x + sin^2x = 1, so you can substitute both cos(t) or sin(t) for x.
 
why is that even possible?

I was watching an MIT Opencourseware lecture and the professor explicitly said that the trig substitutions are tied with x=cos theta and y = sin theta on the unit circle.

The lecture is at:

Thanks for your input.
 
Math_QED said:
Orodruin's reply is a very good one. The only thing you use is the identity cos^2x + sin^2x = 1, so you can substitute both cos(t) or sin(t) for x.

I understand if I substituted x = 1 - sin^2 x because that rests on the pythagorean identity, but that is obviously not the case.
 
RoboNerd said:
why is that even possible?
It is just a substitution of variables, you can use any invertible smooth function for this. The only thing is finding a function so that the resulting integral is simpler. This has nothing to do with polar coordinates or parametrising the unit circle - it is just a variable substitution
 
Do you agree that

$$\int \frac{dx}{\sqrt{1-x^2}} = \int \frac{dy}{\sqrt{1-y^2}}$$
 
we just say in this case that x = y, so yes?
 
but why do they float this idea round? it has to be based off the theory that x = rcos theta
 
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RoboNerd said:
but why do they float this idea round? it has to be based off the theory that x = rcos theta
It has nothing to do with that. It is just a variable substitution.
 
  • #11
So how does this variable substitution work, then?
 
  • #12
The same way as any variable substitution. If you have an integral
$$
\int_a^b f(x) dx
$$
and introduce a one-to-one function ##x = g(t)##, then
$$
\int_a^b f(x) dx = \int_{g^{-1}(a)}^{g^{-1}(b)} f(g(t)) g'(t) dt.
$$
In your case, ##f(x) = 1/\sqrt{1-x^2}## and ##g(t) = \sin(t)##.
 
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thank you for your help!
 

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