Understanding integration with trig identities, and absolute value

[Dethrone]

Homework Statement

In integration, we are allowed to use identities such as sinx = \sqrt{1-cos^2x}. Why does that work, and why doesn't make a difference in integration? Graphing \sqrt{1-cos^2x} is only equal to sinx on certain intervals such as(0, \pi)and (2\pi, 3\pi). More correctly, shouldn't we use the absolute value of \sin\left({x}\right)?

sin^2x = 1 - cos^2x
|sinx| = \sqrt{1 = cos^2x}
or defined piecewisely = {\sin\left({x}\right) in regions where it is above the x-axis, and -\sin\left({x}\right) in regions where x is below the x-axis.

Is it possible to even truly isolate "sin\left({x}\right)" from
sin^2x = 1 - cos^2x? It seems as the |sin\left({x}\right)| is the closest we can to isolate it.

Sorry if I may seem confusing, but the concept of absolute value still confuses me.

Homework Equations

The Attempt at a Solution

 

[1MileCrash]

Can you provide an example? All I feel like I can tell you right now is that you're right, sin(x) is not sqrt(1-cos^2(x)) and that you cannot make that claim in integration either, but I don't know the whole story.

 

[Dethrone]

Well, I was just thinking about this, nothing more. Actually, I don't think this pertains too much to integration, since no one in the right mind would use that substitution. But how would someone isolate sinx from cos^2x + sin^2x = 1, there seems to always be an absolute value in the way. And I really don't think schools teach absolute value well, or at all. I've been taught that the square root of something results in +/-, but it wouldn't make sense that sinx = +/- sqrt(1-cos^2x).

But in generally, in integration, I usually see the books ignoring the absolute value. For example, ∫ x^5\sqrt{1-x^3} dx, you can make the substitution 1 - x^3 = z^2. The square root of z^2 would be the absolute value of z, but they just use integrate z instead.

Also, in ∫\frac{\sqrt{x-x^2}}{x^4} dx, if you use the substitution x = \frac{1}{u}, then you'll end up with -∫ \sqrt{\frac{1}{u^2}(u-1)}u^2 du, and the \frac{1}{u^2} simply becomes \frac{1}{u}. Also, when calculating arc lengths, such as y^3 = 8x^2 from 1 to 8, sqrt(x^(-2/3)) becomes x^(-1/3).

I apologize for the long message, I got addicted to use LaTeX.

 

[Dethrone]

Well, I was just thinking about this, nothing more. Actually, I don't think this pertains too much to integration, since no one in the right mind would use that substitution. But how would someone isolate sinx from cos^2x + sin^2x = 1, there seems to always be an absolute value in the way. And I really don't think schools teach absolute value well, or at all. I've been taught that the square root of something results in +/-, but it wouldn't make sense that sinx = +/- sqrt(1-cos^2x).

But in generally, in integration, I usually see the books ignoring the absolute value. For example, ∫ x^5\sqrt{1-x^3} dx, you can make the substitution 1 - x^3 = z^2. The square root of z^2 would be the absolute value of z, but they just use integrate z instead.

Also, in ∫\frac{\sqrt{x-x^2}}{x^4} dx, if you use the substitution x = \frac{1}{u}, then you'll end up with -∫ \sqrt{\frac{1}{u^2}(u-1)}u^2 du, and the \frac{1}{u^2} simply becomes \frac{1}{u}. Also, when calculating arc lengths, such as y^3 = 8x^2 from 1 to 8, sqrt(x^(-2/3)) becomes x^(-1/3).

I apologize for the long message, I got addicted to use LaTeX.

Anyone have any insights as to why this is true?