An Integral simplification I don't understand

[YouAreAwesome]

Homework Statement

I'm having trouble understanding why the Pythagorean Identity can be applied to this Integration by Substitution.

Relevant Equations

Pythagorean identity

Hi,

I'm missing the jump between these steps.

Why should integrands with different variables be able to use the pythagorean identity? The substitution used was u = 4 - x and the bounds remain the same.

Any help appreciated thanks.

 

[renormalize]

Why should integrands with different variables be able to use the pythagorean identity?
Any help appreciated thanks.

Because $\intop_{a}^{b}f\left(u\right)du$ has exactly the same value as $\intop_{a}^{b}f\left(x\right)dx$. The name of the integration-variable in a definite integral doesn't matter (for that reason, it's called a dummy variable).

 

[YouAreAwesome]

Because $\intop_{a}^{b}f\left(u\right)du$ has exactly the same value as $\intop_{a}^{b}f\left(x\right)dx$. The name of the integration-variable in a definite integral doesn't matter (for that reason, it's called a dummy variable).

Hmmm I understand $\intop_{a}^{b}f\left(u\right)du=\intop_{a}^{b}f\left(x\right)dx$ because each is evaluating the same expression. But how does this relate to numerators of integrands being able to be added together when one is $\cos^2(x)$ and the other is $\sin^2(4-x)$. How do these add to one? It seems to me we are replacing $u$ with $x$ but $u$ does not equal $x$ directly, so I don't see how the pythagorean identity can be used here?

 

[renormalize]

Hmmm I understand $\intop_{a}^{b}f\left(u\right)du=\intop_{a}^{b}f\left(x\right)dx$ because each is evaluating the same expression.

Since you understand this, use that fact to rewrite the second integral and then replace the sum of the two integrals by one integral over the sum of the integrands:$$\intop_{a}^{b}f\left(x\right)dx+\intop_{a}^{b}g\left(u\right)du=\intop_{a}^{b}f\left(x\right)dx+\intop_{a}^{b}g\left(x\right)dx=\intop_{a}^{b}\left[f\left(x\right)+g\left(x\right)\right]dx$$

 

[YouAreAwesome]

Since you understand this, use that fact to rewrite the second integral and then replace the sum of the two integrals by one integral over the sum of the integrands:$$\intop_{a}^{b}f\left(x\right)dx+\intop_{a}^{b}g\left(u\right)du=\intop_{a}^{b}f\left(x\right)dx+\intop_{a}^{b}g\left(x\right)dx=\intop_{a}^{b}\left[f\left(x\right)+g\left(x\right)\right]dx$$

Looking at only the numerators and substituting $u=4-x$ then $$\int_1^3 cos^2(\frac{\pi}{8}x)+sin^2(\frac{\pi}{2}-\frac{\pi}{8}x) dx=2\int_1^3 cos^2(\frac{\pi}{8}x) dx$$

Sorry I'm not seeing why the integrand is equal to 1.

 

[renormalize]

Looking at only the numerators and substituting $u=4-x$

You don't need to do that substitution. After all, you said:

Hmmm I understand $\intop_{a}^{b}f\left(u\right)du=\intop_{a}^{b}f\left(x\right)dx$ because each is evaluating the same expression.

So applied to your specific problem that means:$$\intop_{1}^{3}\frac{\sin^{2}\left(\frac{\pi}{8}u\right)}{u\left(4-u\right)}du=\intop_{1}^{3}\frac{\sin^{2}\left(\frac{\pi}{8}x\right)}{x\left(4-x\right)}dx$$Do you understand?

 

[YouAreAwesome]

Do you understand?

Nope lol.

When I say:

$\intop_{a}^{b}f\left(u\right)du=\intop_{a}^{b}f\left(x\right)dx$

I am meaning I recognize these as equivalent integrals, and therefore they are equal in value.

But it seems to me you are saying this implies $u=x$?
But have we not already established that $u=4-x$?

 

[renormalize]

But it seems to me you are saying this implies $u=x$?

That's what it means for something to be a dummy variable: it doesn't matter whether you call it $u,x,\alpha,\beta$ or anything else!

But have we not already established that $u=4-x$?

I can't say. You never showed how you did the substitution step in your original post.

 

[YouAreAwesome]

That's what it means for something to be a dummy variable: it doesn't matter whether you call it $u,x,\alpha,\beta$ or anything else!

I can't say. You never showed how you did the substitution step in your original post.

Sorry, I did mention it in my original post, it's at the end where I wrote $u=4-x$.

I understand what you mean by 'dummy variable', and that's the reason this is true:

$$\int_{a}^{b}f(u)du=\int_{a}^{b}f(x)dx$$

But we are dealing with this $$\int_{a}^{b}f(u)du=\int_{a}^{b}g(x)dx$$

So I don't see how it's relevant. The first integrand involves $\cos^2x$ while the second involves $\sin^2u$.

 

[renormalize]

I understand what you mean by 'dummy variable', and that's the reason this is true:

$$\int_{a}^{b}f(u)du=\int_{a}^{b}f(x)dx$$

But we are dealing with this $$\int_{a}^{b}f(u)du=\int_{a}^{b}g(x)dx$$

No, we're dealing with your first case. Define $f\left(z\right)=\frac{\sin^{2}\left(\frac{\pi}{8}z\right)}{z\left(4-z\right)}$ and use that $f$ in both sides of the first equation. What do you get?

 

[YouAreAwesome]

No, we're dealing with your first case. Define $f\left(z\right)=\frac{\sin^{2}\left(\frac{\pi}{8}z\right)}{z\left(4-z\right)}$ and use that $f$ in both sides of the first equation. What do you get?

Thanks, you are saying that

$\frac{\sin^{2}\left(\frac{\pi}{8}u\right)}{u\left(4-u\right)}=\frac{\sin^{2}\left(\frac{\pi}{8}x\right)}{x\left(4-x\right)}$

But don't we have to take into account that $u=4-x$?

 

[FactChecker]

But don't we have to take into account that $u=4-x$?

No matter how you arrived at the variable, u, in the OP, it is now a dummy variable for the second integral. You do not show where it came from, but the variable x in $u=4-x$ must have been another dummy variable for the second integral. It is not the same as the dummy variable x in the first integral.

 

[YouAreAwesome]

No matter how you arrived at the variable, u, in the OP, it is now a dummy variable for the second integral. You do not show where it came from, but the variable x in $u=4-x$ must have been another dummy variable for the second integral. It is not the same as the dummy variable x in the first integral.

This is the question I am answering:

Are you saying we can just replace $u$ with $x$ even though $u=4-x$? This is just not clicking for me.

 

[renormalize]

Are you saying we can just replace $u$ with $x$ even though $u=4-x$?

Yes, because it's a dummy variable and so it can have any name. Maybe it's easier to see this way: in the first integral, do the dummy-variable replacement $x\rightarrow v$ and the in the second do $u\rightarrow v$. Now both integrals involve the same new dummy-variable $v$, so you can forget all about $x$ and $u$ and just finish the problem.

 

[YouAreAwesome]

Yes, because it's a dummy variable and so it can have any name. Maybe it's easier to see this way: in the first integral, do the dummy-variable replacement $x\rightarrow v$ and the in the second do $u\rightarrow v$. Now both integrals involve the same new dummy-variable $v$, so you can forget all about $x$ and $u$ and just finish the problem.

Wow, so even though $u=4-x$, we can still use $u=x$ because the area will be same no matter what variable we use for the integral that is currently in terms of $u$?

So is this a true statement then? $$\int_a^b sin^2xdx+\int_a^b cos^2udu= \int_a^b1dx$$

 

[FactChecker]

Wow, so even though $u=4-x$, we can still use $u=x$ because the area will be same no matter what variable we use for the integral that is currently in terms of $u$?

Saying that $u=x$ is sort of misleading. It is just renaming the dummy variable. Suppose you have two integrals, $\int_a^b f(x) dx$ and $\int_c^d g(x) dx$. There is not reason to say that the two dummy variables, $x$, in the two integrals are equal or even related in any way. They are just symbolic dummy variables that go over their respective domains, $[a,b]$ and $[c,d]$.

 

[YouAreAwesome]

@renormalize @FactChecker

Ok I think it just clicked!!! Thank you.

(First I'll double check it did click, but I'm pretty confident it did).

Because we are integrating over a defined domain (the bounds), when we integrate, the variable will be completely removed from the expression -- because we sub in the bounds thus eliminating the 'dummy variable' from the expression altogether. The $x$ and $u$ are completely redundant because of this. It will always just leave us with a numerical value meaning the 'dummy variable' was not important.

So while the pythagorean identity only holds true for equal angles, definite integrals can use the pythagorean identity as long as the bounds are the same in each integral.

So we can literally use $u=x$ in the second integral in the OP even though $u=4-x$ because these variables are not important to the value of the integral.

Is this correct?

 

[SammyS]

@renormalize @FactChecker

Ok I think it just clicked!!! Thank you.

(First I'll double check it did click, but I'm pretty confident it did).

Because we are integrating over a defined domain (the bounds), when we integrate, the variable will be completely removed from the expression -- because we sub in the bounds thus eliminating the 'dummy variable' from the expression altogether. The $x$ and $u$ are completely redundant because of this. It will always just leave us with a numerical value meaning the 'dummy variable' was not important.

In the above paragraph, it's pretty clear that you 'get' the idea of why we call the variable of integration for a definite integral a 'dummy variable'.

So we can literally use $u=x$ in the second integral in the OP even though $u=4-x$ because these variables are not important to the value of the integral.

However, the above sentence (your final sentence) gives me some doubt.

As for using the Pythagorean Identity;
That was used to simplify the resulting integrand after combining two definite integrals. The resulting integral has a relatively easy to find antiderivative. That's not the case for the two initial integrands. The use of the Pythagorean Identity had little or nothing to do with the method of Integration by Substitution.

By the way:
It looks to me like the initial expression in the OP is some intermediate step in some exercise. I may add another post to this thread regarding this as time permits

 

[YouAreAwesome]

However, the above sentence (your final sentence) gives me some doubt.

This is the sentence you are referring to:

So we can literally use u=x in the second integral in the OP even though u=4−x because these variables are not important to the value of the integral.

Ok perhaps if I keep explaining my understanding you might find a mistake along the way that you can correct. The question I'm answering is:

Part ii. We have two equal integrals expressed differently. The given equation has numerator $cos^2(\frac \pi 8 x)$ and the 'shown' equation from Part i. has numerator $sin^2(\frac \pi 8 u)$. Because the dummy variables are inconsequential to the value of the integral, naming them is not important. Important however is that each integral covers the same domain $[1,3]$. Because they cover the same domain, and the variable is not important in a definite integral, and these two equations have equal denominators, we can sum their numerators. Each equation has the same angle with an arbitrary dummy variable, i.e. $\frac \pi 8 *$ (where $*$ is the arbitrary dummy variable). Because they have equal angles we can take their sum.

What do you think? Any errors?

Thanks

 

[SammyS]

Part ii. We have two equal integrals expressed differently. The given equation has numerator $cos^2(\frac \pi 8 x)$ and the shown equations has numerator $sin^2(\frac \pi 8 u)$. Because the dummy variable is inconsequential to the value of the integral, naming them is not important. Important however is that each integral covers the same domain $[1,3]$. Because they cover the same domain, and the variable is not important in a definite integral, these two equations have equal denominators meaning we can sum their numerators. Each equation has the same angle with an arbitrary dummy variable, i.e. $\frac \pi 8 *$ (where $*$ is the arbitrary dummy variable). Because they have equal angles we can take their sum.

What do you think? Any errors?

Thanks

Yes. That all looks good.
Added in an EDIT: I agree with @FactChecker in the following post.
You are missing some important details in justifying some of the most crucial steps.
end of EDIT

I suspected that was the integral you were to evaluate.

Now you have left to evaluate:

$\displaystyle 2I=\int_1^3 \dfrac{1}{x(4-x)}\,dx$

This shouldn't be too difficult.

 

[FactChecker]

Why should integrands with different variables be able to use the pythagorean identity? The substitution used was u = 4 - x and the bounds remain the same.

That is a little misleading. With the change of variable $u=4-x$ you would have $du/dx=-1$ and the integral endpoints would be $\int_3^1$. So several (simple) steps are needed to get the integral in your post. That is not just renaming the dummy variable of an integral.
On the other hand, you could consider the other $u=x$ change of variable to be a simple renaming of the dummy variable or you could say that $du/dx=1$ and consider it a formal change of variable. Either way, these end up as two different dummy variables and how they were arrived at is irrelevant.

 

[SammyS]

That is a little misleading. With the change of variable $u=4-x$ you would have $du/dx=-1$ and the integral endpoints would be $\int_3^1$. So several (simple) steps are needed to get the integral in your post. That is not just renaming the dummy variable of an integral.
On the other hand, you could consider the other $u=x$ change of variable to be a simple renaming of the dummy variable or you could say that $du/dx=1$ and consider it a formal change of variable. Either way, these end up as two different dummy variables and how they were arrived at is irrelevant.

So ...
Seems that you the first to post the issue of the need to change the limits of integration. I'm convinced that much of OP's problem has to do with ignoring that and other details involved with applying Integration by Substitution.

 

[YouAreAwesome]

@SammyS @FactChecker

Again, correct me if I've missed something. But when we use $u=4-x$ the bounds from $x=1$ to $x=3$ become $u=4-1=3$ to $u=4-3=1$ leaving us with $\int_3^1$. We then take the negative of the integral and swap them back to $\int_1^3$.

So several (simple) steps are needed to get the integral in your post.

Perhaps you say this because I privately showed Part i on my own without any issues before posting the relevant question here. To solve Part i, I did as above (let $u=4-x$ etc), then used complementary angles to change '$cos$' to '$sin$' and voilà. Nevertheless, the part that did not click for me was how we can use the pythagorean identity, on what appeared to me to be, two different angles. One $x$ and the other $4-x$. But as you say, what I've learned is, with definite integrals over the same domain, how we get that dummy variable is irrelevant; in other words, it doesn't matter that we used $u=4-x$ to arrive at the integral.

I'm convinced that much of OP's problem has to do with ignoring that and other details involved with applying Integration by Substitution.

Why are you convinced? Is there something I have explained incorrectly or written that reveals a shallow understanding? Which details exactly?

Thank you

 

[SammyS]

@SammyS @FactChecker

Again, correct me if I've missed something. But when we use $u=4-x$ the bounds from $x=1$ to $x=3$ become $u=4-1=3$ to $u=4-3=1$ leaving us with $\int_3^1$. We then take the negative of the integral and swap them back to $\int_1^3$.
...

Thank you

Yes, it's true that taking the negative of the integral will swap the integration bounds. You haven't indicated what justifies taking the negative, but it is the correct thing to do.

It's the lack of mentioning these details which made me of whether you were aware of them.

 

[YouAreAwesome]

Yes, it's true that taking the negative of the integral will swap the integration bounds. You haven't indicated what justifies taking the negative, but it is the correct thing to do.

It's the lack of mentioning these details which made me of whether you were aware of them.

Fair enough. The way I see it, integration can be defined as the (signed) area under a curve. The way we evaluate the total area is by summing the areas of the infinite rectangles between $x=a$ and $x=b$ where each rectangle has height $f(x)$ and width $$\frac{b-a}{n}$$ and $n \rightarrow \infty$. If we swap the bounds we are now summing rectangles with width $$\frac{a-b}{n}$$. But $$b-a=-(a-b)$$. Thus $$\int_a^b f(x)dx=-\int_b^a f(x)dx$$.

Assuming this is correct, if I go through and justify every step, the post will veer off course. My difficulty was solely on how we can sum integrands that involved different variables. I missed that this was a definite integral so the variables, over the same bounds, are arbitrary.

 

[SammyS]

Yes, it's true that taking the negative of the integral will swap the integration bounds. You haven't indicated what justifies taking the negative, but it is the correct thing to do.

It's the lack of mentioning these details which made me of whether you were aware of them.

I'm referring to the specific integral which is the subject of this thread. The integral in posts #13 and #19.
In part $i\,$, when doing Integration by Substitution, where do you get the negative sign which allows swapping the integration limits back to $\displaystyle \int_1^3 \ ?$

 

[YouAreAwesome]

I'm referring to the specific integral which is the subject of this thread. The integral in posts #13 and #19.
In part $i\,$, when doing Integration by Substitution, where do you get the negative sign which allows swapping the integration limits back to $\displaystyle \int_1^3 \ ?$

Let $u=4-x \implies dx=-du$

\begin{align}
\int_1^3 \frac{\cos^2\frac{\pi x}{8}}{x(4-x)}dx & = \int_3^1 \frac{\cos^2\frac{\pi (4-u)}{8}}{(4-u)u} -du \nonumber \\[8pt]
& = -\int_3^1 \frac{\cos^2\left(\frac{4\pi}{8} -\frac{\pi u}{8}\right)}{u(4-u)}du \nonumber \\[8pt]
& = \int_1^3 \frac{\cos^2\left(\frac{\pi}{2} -\frac{\pi u}{8}\right)}{u(4-u)}du \nonumber \\[8pt]
& = \int_1^3 \frac{\sin^2\left(\frac{\pi u}{8}\right)}{u(4-u)}du \nonumber
\end{align}

 

[SammyS]

Perfect !

 

[FactChecker]

@SammyS @FactChecker

Again, correct me if I've missed something. But when we use $u=4-x$ the bounds from $x=1$ to $x=3$ become $u=4-1=3$ to $u=4-3=1$ leaving us with $\int_3^1$. We then take the negative of the integral and swap them back to $\int_1^3$.

Good point. Exactly. So that has a lot more going on than just a symbolic change of a dummy variable. On the other hand, the $u=x$ can be treated as a symbolic change of the dummy variable or as a simple variable substitution with $du/dx = 1$.

Perhaps you say this because I privately showed Part i on my own without any issues before posting the relevant question here. To solve Part i, I did as above (let $u=4-x$ etc), then used complementary angles to change '$cos$' to '$sin$' and voilà. Nevertheless, the part that did not click for me was how we can use the pythagorean identity, on what appeared to me to be, two different angles. One $x$ and the other $4-x$. But as you say, what I've learned is, with definite integrals over the same domain, how we get that dummy variable is irrelevant; in other words, it doesn't matter that we used $u=4-x$ to arrive at the integral.

Exactly. Good work.