- #1

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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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- Thread starter RoboNerd
- Start date

- #1

- 410

- 11

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.

- #2

- 16,923

- 6,730

- #3

member 587159

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.

- #4

- 410

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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.

- #5

- 410

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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.

- #6

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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 substitutionwhy is that even possible?

- #7

- 22,089

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Do you agree that

$$\int \frac{dx}{\sqrt{1-x^2}} = \int \frac{dy}{\sqrt{1-y^2}}$$

$$\int \frac{dx}{\sqrt{1-x^2}} = \int \frac{dy}{\sqrt{1-y^2}}$$

- #8

- 410

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we just say in this case that x = y, so yes?

- #9

- 410

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but why do they float this idea round? it has to be based off the theory that x = rcos theta

- #10

- 16,923

- 6,730

It has nothing to do with that. It is just a variable substitution.but why do they float this idea round? it has to be based off the theory that x = rcos theta

- #11

- 410

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So how does this variable substitution work, then?

- #12

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$$

\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)##.

- #13

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thank you for your help!

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