# SOLVEDIntegration problem

#### matheater

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 thereView attachment 18364 directly.Can anyone help me by showing the procedure step by step?
thanks

Last edited:

#### Also sprach Zarathustra

Use substitution z=p*tan(t).

#### matheater

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?

Last edited by a moderator:

#### 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, $$\displaystyle z'=x$$ and $$\displaystyle \ro = a$$

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

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

so we get:

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

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

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

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

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

Last edited:
• matheater