Integration by substitution (I think)

[tomwilliam]

Homework Statement

Integral of d.cos j with regard to d.sin j

Where d is a constant.

Homework Equations

The Attempt at a Solution

I don't know how to approach this. I can substitute u=d.sin j
Then I have
Integral of dz/dj with regard to dz, but not sure where to go from here.
Any help appreciated.

 

[lanedance]

do you mean a Riemann–Stieltjes integral?
\int f(j) dg(j)

with
\int f(j) = sin(u)
\int f(j) dg(j) = cos(u)

First its probably a bad idea to use d as a symbol for constant in this context, based on its calculus context

Now if f and g has a continuous bounded derivative in a Riemann–Stieltjes integral the following equality holds
\int_a^b f(j) dg(j) = \int_a^b f(j) g'(j) dj

 

[tomwilliam]

Thanks,
Ok, the complexity of your answer tells me I've made an earlier mistake. I was trying to integrate the expression

sqrt(a^2 - b^2)

With regard to b. I used the substitution b=a sin theta so that

Sqrt(a^2(1-sin^2 theta) = sqrt(a^2 cos^2 theta) = a cos theta

Now I have to integrate

a cos theta

With regard to a sin theta. I've changed the variables, but I think that's equivalent to my original post. Did I make a mistake?
Thanks again for your time.

 

[lanedance]

ahh ok so you mean
\int \sqrt{a^2-b^2}db

now let b = a.sin(t)
b = a.sin(t)
db = a.cos(t).dt

subbing in
\int \sqrt{a^2-a^2 sin^2(t)}a.cos(t).dt
\int \sqrt{a^2(1- sin^2(t))}a.cos(t).dt
\int \sqrt{a^2cos^2(t)}a.cos(t).dt

so the integral should be with respect to t (short for theta)

 

[lanedance]

i always do those steps with b & db to keep it clear, similar with limits if the integral has limits

 

[lanedance]

by the way if you ever want to write tex, you can right click on the expression to show source and see how its written

 

[tomwilliam]

Thanks, that's exactly it.
I'm writing on an iPad, which makes it difficult (impossible?) to type tex without putting it into a separate application first.
Thanks for your time.

 

[lanedance]

yeah know the feeling

 

[tomwilliam]

EDIT: It's ok, I've solved it now!