Question

integral \[\int_{0}^{L} \frac{dx}{(z^2+x^2)^{3/2}}\]

Answer 1

\[\int\limits_{0}^{L} {dx\over(z^2+x^2)^{3/2}}\]

Answer 2

please solve it

Answer 3

hm... I can't get it to come up in the equation editor in your post :/
in any event, looks like a trig sub\[x=z\tan\theta\]

Answer 4

pretty sure my sub will get you there

Answer 5

except that answer makes no sense because it still has an x in it

Answer 6

ok try my sub and let me know where you get stuck

Answer 7

i dont no how i m solve it

Answer 8

trig sub\[x=z\tan\theta\]did you try it?

Answer 9

i can not solve it, i m not good in math,, i m physics student and that is use in my problem

Answer 10

can u solve it stepwise

Answer 11

well any good physicist should be able to do a trig sub like this
I have to go to class, so I can't type the whole thing out
feel free to "bump" your Q and let someone else answer it, or learn the technique I'm talking about from here:
http://tutorial.math.lamar.edu/Classes/CalcII/TrigSubstitutions.aspx

Answer 12

you're solving for the electric field above a rod, right?
I think I recognize this equation

Answer 13

yeah you are right

Answer 14

Letting \(x=z \tan\theta,\) we change the bounds accordingly: \[x=0 \Rightarrow \theta=0\\
x=L \Rightarrow \theta=\tan^{-1}\left ( \frac{L}{z}\right )=\alpha\]
Then we deduce that: \(dx=z\ \sec^2\theta\ d\theta\ \)
The integral becomes this:
\[\begin{align}
\large\int_0^L \frac{dx}{(z^2+x^2)^{3/2}}&\large =\int_0^\alpha \frac{z\ \sec^2\theta\ d\theta}{[z^2(1+\tan^2x)]^{3/2}}\\
&\large=\int_0^\alpha \frac{z\ \sec^2\theta\ d\theta}{z^3 \sec^3\theta} \\
&\large= \frac{1}{z^2}\int_0^\alpha \cos\theta\ d\theta\\
&\large=\frac{1}{z^2}\sin\alpha
\end{align}\]
to be continued...

Answer 15

From \(\large \tan\alpha=\frac{L}{z}\), we draw this right triangle to deduce sin(alpha):

From the diagram, we have \(\Large\sin\alpha=\frac{L}{\sqrt{L^2+z^2}}\) so that our final answer is \[\LARGE\frac{L}{z^2(\sqrt{L^2+x^2})}\]
If you have any questions, feel free to ask.