Integration by substitution (I think)

In summary, the integral of a sqrt(a^2 - b^2) with respect to b can be written as \int_a^b \sqrt{a^2-a^2 sin^2(t)}a.cos(t).dt
  • #1
tomwilliam
144
2

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.
 
Physics news on Phys.org
  • #2
do you mean a Riemann–Stieltjes integral?
[tex] \int f(j) dg(j) [/tex]

with
[tex] \int f(j) = sin(u) [/tex]
[tex] \int f(j) dg(j) = cos(u) [/tex]

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
[tex]\int_a^b f(j) dg(j) = \int_a^b f(j) g'(j) dj [/tex]
 
Last edited:
  • #3
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.
 
  • #4
ahh ok so you mean
[tex] \int \sqrt{a^2-b^2}db[/tex]

now let b = a.sin(t)
[tex] b = a.sin(t)[/tex]
[tex] db = a.cos(t).dt[/tex]

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

so the integral should be with respect to t (short for theta)
 
  • #5
i always do those steps with b & db to keep it clear, similar with limits if the integral has limits
 
  • #6
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
 
  • #7
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.
 
  • #8
yeah know the feeling
 
  • #9
EDIT: It's ok, I've solved it now!
 
Last edited:

Related to Integration by substitution (I think)

1. What is integration by substitution?

Integration by substitution is a technique used in calculus to solve integrals. It involves replacing a variable in the integral with a new variable, which simplifies the integral and makes it easier to solve.

2. When should I use integration by substitution?

Integration by substitution is most useful when the integrand (the function inside the integral) contains a composition of functions, such as the chain rule. In these cases, substituting a new variable can make the integral easier to solve.

3. How do I choose the substitution variable?

The best substitution variable to use is usually the innermost function in the integrand. This allows for the simplification of the integral and cancellation of terms in the integrand. However, there is no one right way to choose the substitution variable, and it may take some trial and error to find the best one.

4. What are the steps for integration by substitution?

The steps for integration by substitution are as follows:

  1. Identify the innermost function in the integrand and let u equal this function.
  2. Compute du/dx, the derivative of u with respect to x.
  3. Substitute u and du/dx into the integral, replacing all instances of the innermost function with u and all instances of dx with du.
  4. Integrate the resulting integral with respect to u.
  5. Replace u with the original innermost function to get the final answer.

5. Can integration by substitution be used for all integrals?

No, integration by substitution is not applicable to all integrals. It is most useful for integrals involving compositions of functions, but there are other techniques that may be more appropriate for different types of integrals.

Similar threads

Replies
1
Views
647
  • Calculus and Beyond Homework Help
Replies
2
Views
927
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
Replies
5
Views
862
Replies
4
Views
556
  • Calculus and Beyond Homework Help
Replies
27
Views
3K
  • Calculus and Beyond Homework Help
Replies
12
Views
2K
  • Calculus and Beyond Homework Help
Replies
3
Views
968
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
7
Views
2K
Back
Top