Integration By Substitution It's Been A While

In summary, the student is trying to integrate to find the electric field at a given point. They are confused and stuck on how to solve for du/2x=dx. They use a substitution to find du and get a result of \frac{z^2}{(x^2+z^2)^{3/2}}.
  • #1
Lucretius
152
0

Homework Statement



It's been god knows how long since I've had to use integration by substitution. I've totally forgotten it. I am trying to integrate to solve for the value of an electric field at a given point. The integral I am trying to solve is:

(2kz/4(pi)(epsilon_0)*1/(z^2+x^2)^(3/2) dx.

I know the answer is (2kz/4(pi)(epsilon_0)*(x/[z^2(z^2+x^2)^(1/2)]

Homework Equations



I'm sure I have to set u=(z^2+x^2). This makes du = 2x.

The Attempt at a Solution



I'm confused as to what to do now. The equation I'm integrating doesn't have an x in it anywhere. I don't think I can say du/2x=dx because I will have x's and u's in the integral, which is no good. However, I can't just ignore it.

Also, how did that z^2 get in there on the bottom? z is a constant in this integration and since u = z^2+x^2, the z-term drops right out. I feel terribly lost.
 
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  • #2
That's really confusing. Why don't you just post the actual problem you are working on and how you tried to solve it?
 
  • #3
Sure thing.

I'm trying to find the electric field at an arbitrary distance z above a straight line segment, where the arbitrary distance z is measured above one of the endpoints of the line segment.

Relevant Equations:

We are given that the electric field of a line charge is [tex]\frac{1}{4 \pi \epsilon_0} \int_P \frac{\lambda (R)}{r^2}dl[/tex].

Attempt At Solution.

A little element of the electric field is going to be pointed in two directions. One will be in the z-direction. The other will be in the direction parallel to the line. Using the geometry of the problem, I found that

[tex] dE=\frac{1}{4 \pi \epsilon_0}\frac{\lambda dx}{r^2}(cos(\theta) \textbf{z}+sin(\theta)\textbf{x})=\frac{\lambda}{4 \pi \epsilon_0}\frac{\lamda dx}{r^3}(z\textbf{z}+x\textbf{x})[/tex], where the bold indicates unit vectors.

I split this up into two integrals. This is where I have to integrate by parts, and where I get stuck. I have for instance, one integral which is [tex]\frac{1}{4 \pi \epsilon_0} \int_0^L \frac{2 \lambda z}{(z^2+x^2)^{3/2}}}dx[/tex].

I set my u = (z^2+x^2) and my du is then 2xdx. I am confused because there is no x in my numerator, and I can't just say du/2x=dx because then I am going to have both x's and u's in my equation when I integrate it.
 
  • #4
Lucretius said:
I split this up into two integrals. This is where I have to integrate by parts, and where I get stuck. I have for instance, one integral which is [tex]\frac{1}{4 \pi \epsilon_0} \int_0^L \frac{2 \lambda z}{(z^2+x^2)^{3/2}}}dx[/tex].

I set my u = (z^2+x^2) and my du is then 2xdx. I am confused because there is no x in my numerator, and I can't just say du/2x=dx because then I am going to have both x's and u's in my equation when I integrate it.

Well, if [itex]u = (z^2+x^2)[/itex], then [itex]2x=2\sqrt{u-z^2}[/itex] right?...but I don't think that makes the integral any easier!

Try the substitution [tex]u=\frac{x}{\sqrt{x^2+z^2}}[/tex] instead :wink:
 
  • #5
gabbagabbahey said:
Well, if [itex]u = (z^2+x^2)[/itex], then [itex]2x=2\sqrt{u-z^2}[/itex] right?...but I don't think that makes the integral any easier!

Try the substitution [tex]u=\frac{x}{\sqrt{x^2+z^2}}[/tex] instead :wink:

Ah, I got it, I think.

If I use the u-substitution you suggest, I get [tex]du=\frac{1}{\sqrt{x^2+z^2}}-\frac{x^2}{(x^2+z^2)^{3/2}}dx[/tex], which, getting a common denominator yields:

[tex]\frac{z^2}{(x^2+z^2)^{3/2}}[/tex].

So [tex]\frac{du}{z^2}=\frac{dx}{(x^2+z^2}dx[/tex]

My integral is then just [tex]\frac{\lambda z}{4 \pi \epsilon_0}\int_a^b \frac{du}{z^2}[/tex] which, after re-substituting for u back into x's, and plugging in the bounds 0 and L, I get

[tex] \frac{1}{4 \pi \epsilon_0} \frac{\lambda z L}{z^2(z^2+x^2)^{3/2}}[/tex].

Thanks, though I still wonder how I would have thought of that particular u-substitution on my own!
 

FAQ: Integration By Substitution It's Been A While

What is integration by substitution?

Integration by substitution is a method used in calculus to find the integral of a function by substituting a variable with a new expression. This allows for the integral to be rewritten in a simpler form that can be easily solved.

When should I use integration by substitution?

Integration by substitution is typically used when the integral involves a function within a function, or when it is not immediately obvious how to integrate the function. It can also be used when there is a specific variable that can be substituted to simplify the integral.

What are the steps for integration by substitution?

The general steps for integration by substitution are as follows:
1. Identify a variable to substitute with a new expression.
2. Rewrite the integral in terms of the new variable.
3. Calculate the derivative of the new expression and substitute it back into the integral.
4. Integrate the new integral.
5. Replace the new variable with the original one if necessary.

What are some common mistakes to avoid when using integration by substitution?

Some common mistakes to avoid when using integration by substitution include forgetting to substitute the new variable back into the integral, using the wrong substitution, and not simplifying the integral before integrating. It is also important to carefully check the limits of integration when substituting variables.

Can integration by substitution be used for definite integrals?

Yes, integration by substitution can be used for both indefinite and definite integrals. However, when using it for definite integrals, it is important to pay attention to the limits of integration and make any necessary adjustments to the final answer.

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