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!