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.