# Homework Help: Understanding integration with trig identities, and absolute value

1. Jul 20, 2014

### Dethrone

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

3. The attempt at a solution

2. Jul 20, 2014

### 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.

3. Jul 20, 2014

### 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.

Last edited: Jul 20, 2014
4. Jul 22, 2014

### Dethrone

Anyone have any insights as to why this is true?