# Integrating the delta function

• captainjack2000
This is a great summary. In summary, the student is trying to find the value of the integral between 0 and pi, but they are having trouble figuring out how to proceed. The substitution they are using (u=cosx) results in an equation in which they integrate over the original variable, dx. However, they need this equation to be independent of x, but x=cos^-1(u). Changing the limits of the integral results in the student getting the correct answer.f

## Homework Statement

By using the substitution u=cosx obtain the value of the integral
$$\int$$$$\delta$$(cosx-1/2)dx between 0 and pi

## Homework Equations

I have no idea how to go any further with this apart from substituting in for u!?

## The Attempt at a Solution

What's the integral you get after doing the substitution?

So, make the substitution! We can't help you if you don't do (or show) your work.

Hi captainjack2000!

(have a delta: δ )
By using the substitution u=cosx obtain the value of the integral
$$\int$$$$\delta$$(cosx-1/2)dx between 0 and pi

Hint: what is the definition of the Dirac delta function?

Sorry, I realize my post was very specific. To be perfectly honest I don't really know where to start. I know that the delta function is defined as
$$\delta$$(x-a)=o if x$$\neq$$a and tends to infinity is x=a
shifting property
$$\int$$dx$$\delta$$(x-a)f(x) = f(a) where the integral is over all space.

In this case
I=$$\int$$$$\delta$$(cosx-0.5)dx where the integral is between zero and pi
=$$\int$$$$\delta$$(u-0.5)dx letting u=cosx

not sure how to proceed

You're substituting, yet after the substitution you integrate over the original variable,dx? Execute the substitution correctly first. Also put the entire expression between $$brackets it saves you time and is a lot neater. You still have dx in your integral. Write everything in terms of u and du. Don't forget to change the limits of the integral too. Ok...(my brain is obviously not working right now!) if u=cosx du=-sinxdx dx=-du/sinx but you need this independent of x? but x=cos^-1(u) ?? changing the limits: when x=0 u=1 and when x=pi u=-1 You can write sin x = √(1 - u2). So the integral becomes [tex] -\int_{1}^{-1} \delta(u - 1/2) \frac{1}{\sqrt{1-u^2}}du$$

To see that $\sin \arccos u =\sqrt{1-u^2}$ use $\cos^2x+\sin^2x=1$ or use a triangle.

Last edited:
To see that $\sin \arccos u =\frac{1}{\sqrt{1-u^2}}$ use $\cos^2x+\sin^2x=1$ or use a triangle.

You mean

$$\sin \cos^{-1} u = \sqrt{1-u^2}$$.

Yes I do thanks for pointing it out, it's fixed now.

thank you that clarifies it alot!
Therefore the solution is going to be -1/(1-1/4)^0.5 = 1/(3/4)^0.5 = 2/sqrt(3) ?

Yep, that's correct.