Variable substititon

[Arkimedes]

When integrating

$\displaystyle \int_{0}^{\frac{L}{4}}sin^2(\pi \frac{x}{L}) dx$

And making the variable substitution $\displaystyle \theta=\pi \frac{x}{L}$ we get the integral

$\displaystyle \int_{0}^{\frac{\pi}{4}}sin^2(\theta) d\theta$

But I don't understand why $\displaystyle \frac{L}{4}$ becomes $\displaystyle \frac{\pi}{4}$ ?

 

[Deleted member 4993]

When integrating

$\displaystyle \int_{0}^{\frac{L}{4}}sin^2(\pi \frac{x}{L}) dx$

And making the variable substitution $\displaystyle \theta=\pi \frac{x}{L}$ we get the integral

$\displaystyle \int_{0}^{\frac{\pi}{4}}sin^2(\theta) d\theta$

But I don't understand why $\displaystyle \frac{L}{4}$ becomes $\displaystyle \frac{\pi}{4}$ ?

According to the substitution

when x = L/4

$\displaystyle \theta=\pi \dfrac{\frac{L}{4}}{L}$

 

[Arkimedes]

According to the substitution

when x = L/4

$\displaystyle \theta=\pi \dfrac{\frac{L}{4}}{L}$

Thank you friend

 

[Arkimedes]

Ok great! I then just wonder, if we add $\displaystyle \frac{2}{L}$ to the integral, so it becomes:

$\displaystyle \int_{0}^{\frac{L}{4}}\frac{2}{L}sin^2(\pi \frac{x}{L}) dx$

And making the same variable substitution, how do you know how to substitute the the expression $\displaystyle \frac{2}{L}$ ?

 

[Steven G]

When integrating

$\displaystyle \int_{0}^{\frac{L}{4}}sin^2(\pi \frac{x}{L}) dx$

And making the variable substitution $\displaystyle \theta=\pi \frac{x}{L}$ we get the integral

$\displaystyle \int_{0}^{\frac{\pi}{4}}sin^2(\theta) d\theta$

But I don't understand why $\displaystyle \frac{L}{4}$ becomes $\displaystyle \frac{\pi}{4}$ ?

You also do not understand why the lower limit 0 becomes 0

$\displaystyle \theta=\pi \frac{x}{L}$ so instead of the angle being $\displaystyle \pi \frac{x}{L}$ it is just $\displaystyle \theta$. But what about dx? Why does it become d$\displaystyle \theta$???? It is true that d$\displaystyle \theta$=$\displaystyle \frac{pi}{L}$dx. Now solve this for dx.

The answer to your question: when x = $\displaystyle \frac{L}{4}$, $\displaystyle \theta$ = $\displaystyle \frac{\pi}{4}$ AND when x=0, $\displaystyle \theta$ = 0

 

[Deleted member 4993]

Ok great! I then just wonder, if we add $\displaystyle \frac{2}{L}$ to the integral, so it becomes:

$\displaystyle \int_{0}^{\frac{L}{4}}\frac{2}{L}sin^2(\pi \frac{x}{L}) dx$

And making the same variable substitution, how do you know how to substitute the the expression $\displaystyle \frac{2}{L}$ ?

$\displaystyle \frac{2}{L}$ is not a function of 'x'. So it is not affected by the substitution for 'x'. So $\displaystyle \frac{2}{L}$ remains as $\displaystyle \frac{2}{L}$.

 

[Arkimedes]

$\displaystyle \frac{2}{L}$ is not a function of 'x'. So it is not affected by the substitution for 'x'. So $\displaystyle \frac{2}{L}$ remains as $\displaystyle \frac{2}{L}$.

What you are saying sounds like it's making sense to me but I actually took this example from a textbook. And there they also change $\displaystyle \frac{2}{L}$, it doesn't remain the same. I copied the steps in the image below:



I don't understand why $\displaystyle \frac{2}{L}$ becomes $\displaystyle \frac{2}{\pi}$. Please explain why this happens???

 

[Arkimedes]

This example is actually from my textbook:



As you can see, for some reason 2/L became 2/pi. I understand why the integration interval changes now but Why does 2/L change to 2/pi ???

 

[Deleted member 4993]

This example is actually from my textbook:

View attachment 10553

As you can see, for some reason 2/L became 2/pi. I understand why the integration interval changes now but Why does 2/L change to 2/pi ???

You need to work with the expression altogether!!!

Substitute:

$\displaystyle \theta = \frac{\pi x}{L}$

$\displaystyle d\theta = \frac{\pi}{L}dx$

$\displaystyle dx = \frac{L}{\pi}d\theta$

Then

$\displaystyle \displaystyle{\frac{2}{L}sin^2\frac{\pi x}{L}dx}$

$\displaystyle \displaystyle{= \frac{2}{L}sin^2\theta dx}$ ........ now substitute for "dx"

$\displaystyle \displaystyle{= \frac{2}{L}sin^2\theta [\frac{L}{\pi}d\theta}]$

$\displaystyle \displaystyle{= \frac{2}{L}\frac{L}{\pi}sin^2\theta [d\theta}]$ .... continue....

 

[Arkimedes]

Finally, I get it now. Thanks a lot!