Question on basic trig substitution with x = sin theta

• I
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
Staff Emeritus
Homework Helper
Gold Member
2021 Award
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.

member 587159
member 587159
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:

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.

Orodruin
Staff Emeritus
Homework Helper
Gold Member
2021 Award
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

micromass
Staff Emeritus
Homework Helper
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

Orodruin
Staff Emeritus
Homework Helper
Gold Member
2021 Award
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.

So how does this variable substitution work, then?

Orodruin
Staff Emeritus
$$\int_a^b f(x) dx$$
$$\int_a^b f(x) dx = \int_{g^{-1}(a)}^{g^{-1}(b)} f(g(t)) g'(t) dt.$$