Oopsie, issue with change of variables to evaluate definite integral

[BobbyBear]

I needed to evaluate the following integral (for constructing Chebyshev polynomials by an orthogonalisation process), but I just discovered that I'm having an issue with the change of variable technique:P The specific integral itself is unimportant as to the issue I'm having, but by means of an example this is the one I was solving when it arose:

<br /> \int_{-1}^{1} \frac{1}{\sqrt{1 - x^2}} dx<br />

And I believe this is done with the following change of variable:

<br /> x = g(\theta) = cos (\theta); \ \ \rightarrow \ \ dx= g&#039;(\theta) = -sin(\theta)d\theta<br />

But since the transformation has to be bijective (or however you say:P), one would have to limit the range of \theta otherwise the inverse relationship \theta = g^{-1}(x) would not be a single valued function.

So for example, I can restrict \theta as so: <br /> 0 \leq\theta \leq \pi<br />
in which case g^{-1}(x=-1)=\pi ; \ \ \ g^{-1}(x=1)=0
and thus:
<br /> \int_{-1}^{1} \frac{1}{\sqrt{1 - x^2}} dx = \int_{\pi}^{0} \frac{-sin(\theta)}{\sqrt{(sin(\theta)^2}} d\theta = \int_{\pi}^{0} -d\theta = \int_{0}^{\pi} d\theta = \pi<br />

BUT! if for example I use the same transformation but restricting \theta to: <br /> \pi \leq\theta \leq 2\pi<br />
then I have: g^{-1}(x=-1)=\pi ; \ \ \ g^{-1}(x=1)=2\pi
and:
<br /> \int_{-1}^{1} \frac{1}{\sqrt{1 - x^2}} dx = \int_{\pi}^{2\pi} \frac{-sin(\theta)}{\sqrt{(sin(\theta)^2}} d\theta = \int_{\pi}^{2\pi} -d\theta = -\pi<br />

I know the answer has to be \pi because the integrand in terms of the original variable is always positive for x\in (-1,1), thus the integral has to be positive. And I know that different signs I'm getting is because in one case my transformation is such that \theta is increasing with increasing x (ie. g&#039;&gt;0) and in the other \theta is decreasing with increasing x (ie. g&#039;&lt;0), but . . . the sign of g&#039;&lt;0 should compensate the change in the order of the limits of integration and I shouldn't have to worry about that, no? I just can't see what I'm doing wrong in either case, although I'm certainly doing something wrong!:(

 

[Chaos2009]

I'm not sure if this helps but I recognize that:
\frac{d}{dx}(arcsin(x)) = \frac{1}{\sqrt{1 - x^2}}
where -\frac{\pi}{2} &lt; x &lt; \frac{\pi}{2}

 

[sutupidmath]

People always forget that:

\sqrt{x^2}=|x| and not simply x.

Frankly, i didn't bother to look your whole work, but i think that your problem is, as i just stated:


\sqrt{sin^2(\theta)}=|sin(\theta)|

and not as you, somewhere along the lines, have done simply sin(@) (@=theta)


Now, i assume you know the definition of the abs value, right? Well, in case you cannot recall while reading this post:

|x|=x if x>0 and -x if x<0.

Now:

|sin(@)|=sin(@) if sin(@)<0 and -sin(@) if sin(@)>0

Look at your limits of integration, pi to 2pi. what do you notice? THis is your problem..

 

[BobbyBear]

sutupidmath, . . . yes!

People always forget that:
\sqrt{x^2}=|x| and not simply x.

that includes me

My problem was not with the change of variable theorem after all :P Now everything falls into place :)

Fanku, fanku, I am much indebted :P

 

[BobbyBear]

Thanku to Chaos2009 as well for their observation:P

 

[sutupidmath]

|sin(@)|=sin(@) if sin(@)<0 and -sin(@) if sin(@)>0

..

this should have read |sin(@)|=sin@ if sin@>0 and -sin@ if sin@<0
My bad..

 

[BobbyBear]

Lol that's okay, actually, I got your meaning without even realising you said it the wrong way round:P