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Integration Method for Irrational Root?

  • Thread starter Seydlitz
  • Start date
  • #1
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Homework Statement



I need to evaluate this particular integral and I'm confused on what method to use. I'm currently learning integration calculus and I tried doing some introduction on electromagnetic field. Quite unexpectedly the integral turned to be heavy.

[tex]\int_{-a}^a \frac{1}{\left(x^2+y^2\right)^{3/2}} \, dy[/tex]

The Attempt at a Solution



I have tried on using integration by substitution. I came up with this indefinite integral which is not correct according to the solution:

Let u = [tex]\left(x^2+y^2\right)[/tex]
du = [tex]2y\text{dy}[/tex]
Hence:
[tex]\int \frac{1}{\left(x^2+y^2\right)^{3/2}} \, dy
\int \frac{u^{-3/2}}{2} \, du
\frac{1}{2} \int u^{-3/2} \, du
\frac{\frac{1}{\sqrt{u}}}{\frac{2 (-1)}{2}}-\frac{1}{\sqrt{u}}-\frac{1}{\sqrt{x^2+y^2}}-\frac{1}{\sqrt{x^2+y^2}}[/tex]
[tex]-\frac{1}{\sqrt{x^2+y^2}}[/tex]

If it is possible I also would like to know what type of integral is this because I'm afraid I've not reached the level for this type of problem. I'm sorry if my formatting is bad, this is my first time using LaTex.

Thank You
 
Last edited:

Answers and Replies

  • #2
vela
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Homework Statement



I need to evaluate this particular integral and I'm confused on what method to use. I'm currently learning integration calculus and I tried doing some introduction on electromagnetic field. Quite unexpectedly the integral turned to be heavy.

[tex]\int_{-a}^a \frac{1}{\left(x^2+y^2\right)^{3/2}} \, dy[/tex]

The Attempt at a Solution



I have tried on using integration by substitution. I came up with this indefinite integral which is not correct according to the solution:

Let u = [tex]\left(x^2+y^2\right)[/tex]
du = [tex]2y\text{dy}[/tex]
With this substution, you'd get dy = du/2y and [itex]y=\sqrt{u-x^2}[/itex], so the integral would become
[tex]\int \frac{1}{2u^{3/2}\sqrt{u-x^2}}\,du[/tex]so you wouldn't get
Hence:
[tex]-\frac{1}{\sqrt{x^2+y^2}}[/tex]
Try the substitution [itex]y = x \tan \theta[/itex].
 
  • #3
263
4
I've added my complete working in the first post as might be required.

I'm sorry but how could you determine to substitute y = x tan θ? I know that the expression is equal to y itself, but how to do the integration with the trigonometric function θ?
 
  • #4
HallsofIvy
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That's called "trigonometric substitution". Whenever you have something like "[itex]a^2+ x^2[/itex]", :[itex]a^2- x^2[/itex]", or "[itex]x^2- a^2[/itex]" inside a root you should think of [itex]sin^2(\theta)+ cos^2(\theta)= 1[/itex] and its variations: [itex]1+ tan^2(\theta)= sec^2(\theta)[/itex], [itex]1- sin^2(\theta)= cos^2(\theta)[/itex], and [itex]sec^2(\theta)- 1= tan^2(\theta)[/itex]. Setting [itex]x= atan(\theta)[/itex], [itex]x= asin(\theta)[/itex], and [itex]x= asec(\theta)[/itex] converts each of the terms in x, above, into the corresponding trig formula and so gives a perfect square inside the root.

Since the hyperbolic functions satisfy similar identities: [itex]cosh^2(x)- sinh^2(x)= 1[/itex], etc., they can also be used.
 
  • #5
263
4
Ah thank you, I'll try so solve the problem and I'll let you know the result.
 

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