# Integrating the delta function

## 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!?

## Answers and Replies

dx
Homework Helper
Gold Member
What's the integral you get after doing the substitution?

D H
Staff Emeritus
Science Advisor
So, make the substitution! We can't help you if you don't do (or show) your work.

tiny-tim
Science Advisor
Homework Helper
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 realise 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

Cyosis
Homework Helper
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 $$brackets it saves you time and is a lot neater. dx Homework Helper Gold Member You still have dx in your integral. Write everything in terms of u and du. Don't forget to change the limits of the integral too. Ok...(my brain is obviously not working right now!) if u=cosx du=-sinxdx dx=-du/sinx but you need this independent of x? but x=cos^-1(u) ?? changing the limits: when x=0 u=1 and when x=pi u=-1 dx Homework Helper Gold Member You can write sin x = √(1 - u2). So the integral becomes [tex] -\int_{1}^{-1} \delta(u - 1/2) \frac{1}{\sqrt{1-u^2}}du$$

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

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

You mean

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

Cyosis
Homework Helper
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) ?

Cyosis
Homework Helper
Yep, that's correct.

dx
Homework Helper
Gold Member
Your final answer is correct, but you had a minus sign in your first expression which shouldn't be there.

thank you !