1. ## Integration problem

hi all
I am a new member here.Now I want to go to my problem directly,the problem is about an integration,i didn't understand it clearly.Here i have attached the document.It is done thereAttachment 18364 directly.Can anyone help me by showing the procedure step by step?
thanks

2. Use substitution z=p*tan(t).

3. thanks a lot, can u help more elaborately? I am still unable to find how they got H=I/(2pi.roh)

if we use tan(t) then the limit will be changed to -90 to +90 degrees isn't it? then how can we proceed?

4. Originally Posted by Also sprach Zarathustra
Use substitution z=p*tan(t).

Step by step solution:

First of all I will change the your variables(for my own comfort)

So, $z'=x$ and $\ro = a$

Now, we use following substitution $x=atan(t)$, we "put" it in:

$\frac{1}{{(a^2+x^2)}^{\frac{3}{2}}}$

so we get:

$\frac{1}{{{(a^2+a^2tan^2t)}}^{\frac{3}{2}}}=\frac{ 1}{{{(a^2(1+tan^2(t))}}^{\frac{3}{2}}}=\frac{1}{{{ (a^2(sec^2(t))}}^{\frac{3}{2}}}=\frac{1}{a^3sec^3( t)}$

and
$dx=asec^2(t)dt$

so, $\int\frac{1}{{{(a^2+x^2)}}^{\frac{3}{2}}}dx=\frac{ 1}{a}\int cos(t)dt=\frac{sin(t)}{a}+C$

Now, to return back to $x$, substitute $t=arctan(\frac{x}{a})$

I will leave the last part for you, only trigonometric calculation.