Question

How would you integarte this:

Answer 1

\[\int\limits dz /(r ^{2}+z ^{2})^{3/2}\]

Answer 2

is z complex no?

Answer 3

no

Answer 4

try usual trick ... trig substitution ... z = r tan theta

Answer 5

r^(2) is just a constant remember

try to get 1/(r2+z2)^3/2

you need it in the form

1/(1 + z^(2))^(1/2)
then you can use trigonometric substitution to solve

so lets factor

\[\frac{1}{(r^{6})^{1/2}(1+z^{6})^{1/2}}\]

Answer 6

do you want it as series?

Answer 7

crud

Answer 8

no

Answer 9

it should be

\[\frac{1}{(r^{6})^{1/2}(1 + \frac{z^{6}}{r^{2}})^{1/2}}\]

Answer 10

this doesn't look good ... from other method

Answer 11

so you have the integral:
\[\frac{1}{(r^{6})^{1/2}}\int\limits_{}^{}\frac{1}{(1 + \frac{z^{6}}{r^{2}})^{1/2}}\]

now use trigonometric substation to solve:
here is a video that shows you how to do trig sub:
http://www.youtube.com/watch?v=n4EK92CSuBE

this is the list of the three methods and what trig identity to use

http://en.wikipedia.org/wiki/Trigonometric_substitution

Answer 12

can you do it in a *tan t method? what do you get for x^2+a^2?

Answer 13

in your case you would use tan(x)

Answer 14

let me verify