MHB Integration with trig identities and absolute value

[Dethrone]

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.

 

[I like Serena]

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.

Hi Rido12! :)

As you surmise, we cannot just use $$\sin x = \sqrt{1-\cos^2x}$$.
Depending on the value of x, it can also be $$\sin x = -\sqrt{1-\cos^2x}$$

If we use the expression in an integral, we need to split the integral in separate integrals and use the appropriate version in each integral.

For instance:
$$\int_0^{2\pi} \sin x \, dx
= \int_0^{\pi} \sin x \, dx + \int_{\pi}^{2\pi} \sin x \, dx
= \int_0^{\pi} \sqrt{1-\cos^2x} \, dx + \int_{\pi}^{2\pi} -\sqrt{1-\cos^2x} \, dx$$

 

[Dethrone]

Hi Rido12! :)

As you surmise, we cannot just use $$\sin x = \sqrt{1-\cos^2x}$$.
Depending on the value of x, it can also be $$\sin x = -\sqrt{1-\cos^2x}$$

If we use the expression in an integral, we need to split the integral in separate integrals and use the appropriate version in each integral.

For instance:
$$\int_0^{2\pi} \sin x \, dx
= \int_0^{\pi} \sin x \, dx + \int_{\pi}^{2\pi} \sin x \, dx
= \int_0^{\pi} \sqrt{1-\cos^2x} \, dx + \int_{\pi}^{2\pi} -\sqrt{1-\cos^2x} \, dx$$

Thank you! My suspicions were correct. Btw, do you have any ideas as to why the absolute value is generally left out / ignored in integration, as in the following integral?

$$∫ x^5\sqrt{1-x^3} dx$$ , you can make the substitution $$ 1 - x^3 = z^2$$. The square root of $$z^2 $$ should then become |z|, but instead, it is just left as "z".

 

[I like Serena]

Thank you! My suspicions were correct. Btw, do you have any ideas as to why the absolute value is generally left out / ignored in integration, as in the following integral?

$$∫ x^5\sqrt{1-x^3} dx$$ , you can make the substitution $$ 1 - x^3 = z^2$$. The square root of $$z^2 $$ should then become |z|, but instead, it is just left as "z".

It's a choice.

We can pick $z$ either positive or negative, but not both.
The reason is that $1-x^3$ is a function that is monotone between 0 and infinity.
That means we have to pick $z$ such that $z^2$ is also monotone between 0 and infinity.
That is only the case if we either pick $z$ to be negative or we pick $z$ to be positive.

Calculating with a positive $z$ is usually easier, so that choice is often implicitly made.

 

[Dethrone]

What do you mean by "monotone"? You mean that it's like an even function? And that the interval (0, infinity) is the same as (-infinity, 0)?

 

[Prove It]

Thank you! My suspicions were correct. Btw, do you have any ideas as to why the absolute value is generally left out / ignored in integration, as in the following integral?

$$∫ x^5\sqrt{1-x^3} dx$$ , you can make the substitution $$ 1 - x^3 = z^2$$. The square root of $$z^2 $$ should then become |z|, but instead, it is just left as "z".

While you probably could use that substitution, a better one (due to lack of ambiguity) is the substitution $\displaystyle \begin{align*} z = 1 - x^3 \implies \mathrm{d}z = -3x^2 \end{align*}$, giving

$\displaystyle \begin{align*} \int{ x^5 \, \sqrt{1 - x^3}\,\mathrm{d}x } &= - \frac{1}{3} \int{ x^3 \,\sqrt{1 - x^3} \, \left( -3x^2 \right) \, \mathrm{d}x } \\ &= -\frac{1}{3}\int{ \left( 1 - z \right) \, \sqrt{ z } \, \mathrm{d}z } \\ &= -\frac{1}{3} \int{ z^{\frac{1}{2}} - z^{\frac{3}{2}}\,\mathrm{d}z} \end{align*}$

and I'm sure you can go from here without any worry :)

 

[I like Serena]

What do you mean by "monotone"? You mean that it's like an even function? And that the interval (0, infinity) is the same as (-infinity, 0)?

Monotone means that the function passes the horizontal line test, which is pretty much required to switch to a different variable.

Both $1-x^3$ and $z^2$ have to pass this horizontal line test.
Furthermore, both need to be non-negative, since we need to be able to put them under a square root.

It means the $x$ is restricted to $(-\infty, 1]$ and that $z$ is restricted to either $(-\infty, 0]$ or $[0,\infty)$.To make it more explicit, if we evaluate the definite integral:
$$\int_{-2}^{1} x^5\sqrt{1-x^3} dx$$
We would get:
$$1-x^3=z^2 \quad \Rightarrow\quad x = (1-z^2)^{1/3} \quad \Rightarrow \quad dx = -\frac 2 3 z(1-z^2)^{-2/3} dz$$
If we pick $z \ge 0$ this becomes:
$$\int_{-2}^{1} x^5\sqrt{1-x^3} dx = \int_3^0 (1-z^2)^{5/3} \sqrt{z^2} \cdot -\frac 2 3 z(1-z^2)^{-2/3} dz = -\frac 2 3 \int_3^0 z(1-z^2) \sqrt{z^2}dz$$
Note that the direction of integration flips (from 3 down to 0) and also that a minus sign is introduced that effectively cancels that.

Since $z \ge 0$, we can now replace $\sqrt{z^2}$ by $z$ and we get:
$$-\frac 2 3 \int_3^0 z(1-z^2) \sqrt{z^2}dz = -\frac 2 3 \int_3^0 z(1-z^2) \cdot z\, dz$$

 

[Dethrone]

I understand all of it, except,

Note that the direction of integration flips (from 3 down to 0) and also that a minus sign is introduced that effectively cancels that.

Are you referring to if $$z \le 0$$, then the direction of integration flips as well as an extra minus sign that cancels with it?

 

[I like Serena]

I understand all of it, except,Are you referring to if $$z \le 0$$, then the direction of integration flips as well as an extra minus sign that cancels with it?

If we pick $z\le 0$ the direction of integration does not flip.
It will be from -3 up to 0.

 

[Dethrone]

You said above that if $$z \ge 0$$, then
$$ -\frac 2 3 \int_3^0 z(1-z^2) \sqrt{z^2}dz = -\frac 2 3 \int_3^0 z(1-z^2) \cdot z\, dz$$

Then if $$z \le 0$$, wouldn't it be
$$ -\frac 2 3 \int_3^0 z(1-z^2) \sqrt{z^2}dz = -\frac 2 3 \int_3^0 z(1-z^2) \cdot -z\, dz $$ = $$\frac{2}{3} \int_3^0 z^2 (1-z^2)$$

Don't the two represent different areas? With the top being a positive quantity and the bottom being a negative?

By the way, I really appreciate the help! I love this forum, everyone here is incredibly helpful :)

 

[I like Serena]

You said above that if $$z \ge 0$$, then
$$ -\frac 2 3 \int_3^0 z(1-z^2) \sqrt{z^2}dz = -\frac 2 3 \int_3^0 z(1-z^2) \cdot z\, dz$$

Then if $$z \le 0$$, wouldn't it be
$$ -\frac 2 3 \int_3^0 z(1-z^2) \sqrt{z^2}dz = -\frac 2 3 \int_3^0 z(1-z^2) \cdot -z\, dz $$ = $$\frac{2}{3} \int_3^0 z^2 (1-z^2)$$

Don't the two represent different areas? With the top being a positive quantity and the bottom being a negative?

With $z \le 0$, the lower boundary should be -3 instead of 3.
Combined with the fact that $\int_a^b f(x)dx = -\int_b^a f(x)dx$ they come out the same.

By the way, I really appreciate the help! I love this forum, everyone here is incredibly helpful :)

Thanks! ;)

 

[I like Serena]

What do you mean by "monotone"? You mean that it's like an even function? And that the interval (0, infinity) is the same as (-infinity, 0)?

To clarify, from wikipedia, the rule for substitution is:

Let $I ⊆ ℝ$ be an interval and $\phi: [a,b] → I$ be a continuously differentiable function.
Suppose that $f : I → ℝ$ is a continuous function. Then:
$$
\int_{\phi(a)}^{\phi(b)} f(x)\,dx = \int_a^b f(\phi(t))\phi'(t)\, dt.
$$
​

In your case, you want to replace $1-x^3$ by $z^2$.
Or more specifically, replace $\sqrt{1-x^3}$ by either $z$ or by $-z$.
Now the function $\phi$ is supposed to map $z$ to $x$.
For that to happen, we need that the function $x \mapsto \sqrt{1-x^3}$ is invertible and the inverse function must be continuously differentiable.
A typical way to achieve that is when the function is strictly monotone and differentiable.

It means that you would pick for instance:
$$\phi: z \mapsto (1-z^2)^{1/3}$$
and
$$f: x \mapsto x^5\sqrt{1-x^3}$$
after which you can apply the substitution rule as mentioned on wiki.