# I Question on basic trig substitution with x = sin theta

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1. May 29, 2016

### RoboNerd

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.

2. May 29, 2016

### Orodruin

Staff Emeritus
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.

3. May 29, 2016

### Math_QED

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. May 29, 2016

### RoboNerd

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:

5. May 29, 2016

### RoboNerd

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. May 29, 2016

### Orodruin

Staff Emeritus
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

7. May 29, 2016

### micromass

Staff Emeritus
Do you agree that

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

8. May 30, 2016

### RoboNerd

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

9. May 30, 2016

### RoboNerd

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

10. May 30, 2016

### Orodruin

Staff Emeritus
It has nothing to do with that. It is just a variable substitution.

11. May 30, 2016

### RoboNerd

So how does this variable substitution work, then?

12. May 30, 2016

### Orodruin

Staff Emeritus
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 $g(t) = \sin(t)$.

13. Jun 1, 2016