SOLVED Integration problem

Jul 2010
2
0
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
 

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Jul 2010
2
0
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?
 
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Dec 2009
1,506
434
Russia
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.
 
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