- #1
stevendaryl said:We're trying to integrate
[itex]\frac{z-Ru}{(R^2 + z^2 - 2Rzu)^{\frac{3}{2}}}[/itex]
We can rewrite this as:
[itex]\frac{R}{(2Rz)^{\frac{3}{2}}} \frac{\frac{z}{R} - u}{(\frac{R^2 + z^2}{2Rz} - u)^\frac{3}{2}}[/itex]
Now [itex]\frac{\frac{z}{R} - u}{(\frac{R^2 + z^2}{2Rz} - u)^\frac{3}{2}}[/itex] is of the form
[itex](A-u)(C-u)^\alpha[/itex]
with [itex]A = \frac{z}{R}[/itex], [itex]C = \frac{R^2 + z^2}{2Rz}[/itex], [itex]\alpha = \frac{-3}{2}[/itex]
Here's the general way to integrate such expressions:
Rewrite this as:
[itex]((A-C) + (C-u))(C-u)^\alpha = (A-C) (C-u)^\alpha + (C-u)^{\alpha+1}[/itex]
We can easily integrate these terms, to get:
[itex]-\frac{A-C}{\alpha+1} (C-u)^{\alpha+1} - \frac{1}{\alpha+2} (C-u)^{\alpha+2}[/itex]
[itex]= -\frac{(C-u)^{\alpha+1}}{(\alpha+1)(\alpha+2)} ((A-C)(\alpha+2) + (C-u)(\alpha+1))[/itex]
[itex]= -\frac{(C-u)^{\alpha+1}}{(\alpha+1)(\alpha+2)} (A (\alpha+2) - C - u(\alpha+1))[/itex]
So if you plug in our known values for [itex]A, C, \alpha[/itex], you should get the desired result (after some simplification).
garylau said:where is this form
[itex](A-u)(C-u)^\alpha[/itex]
coming from
I suspect both of those relationships came from a book of integrals.garylau said:where is this form
[itex](A-u)(C-u)^\alpha[/itex]
coming from
and Where do you got [itex]((A-C) + (C-u))(C-u)^\alpha = (A-C) (C-u)^\alpha + (C-u)^{\alpha+1}[/itex] this equation from?
thank
garylau said:where is this form
[itex](A-u)(C-u)^\alpha[/itex]
coming from
...and Where do you got [itex]((A-C) + (C-u))(C-u)^\alpha = (A-C) (C-u)^\alpha + (C-u)^{\alpha+1}[/itex] this equation from?
Ye gads, is this not cheating? Would that actually work, and if so is it easier?Dragon27 said:But the temptation to integrate with respect to ##z## (since ##(z^2-Rzu)'=2(z-Ru)##), then integrate with respect to ##u##, and then differentiate with respect to ##z## is just irresistible...
It feels so good, that it probably is...Battlemage! said:Ye gads, is this not cheating?
The integrals in the intermediate steps are much easier, in my opinion (or, maybe, more obvious). In the end you just differentiate a simple fraction, it's not too hard.Battlemage! said:Would that actually work, and if so is it easier?
Integrating a function is a mathematical process used to find the area under a curve between two given points. It is often used in physics, engineering, and other scientific fields to calculate things such as displacement, velocity, and acceleration.
The method of integration used depends on the type of function being integrated. Some common methods include substitution, integration by parts, and trigonometric substitution. It is important to understand the properties of each method and determine which one is best suited for the given function.
Yes, there are many online calculators and software programs that can integrate functions for you. However, it is important to have a solid understanding of the integration process and the ability to check the accuracy of the results.
The basic steps for integrating a function include identifying the limits of integration, choosing an appropriate method, integrating the function, and evaluating the definite integral using the limits of integration. It is also important to simplify the result and check for any errors.
Yes, there are some special cases and exceptions when integrating a function. These include functions with vertical asymptotes, discontinuities, and improper integrals. It is important to be aware of these cases and use proper techniques to integrate them.