Integration Method for Irrational Root?

In summary, the student is trying to solve an indefinite integral and is confused on what method to use. He tried using integration by substitution and ended up getting the wrong answer. He then tried using the hyperbolic functions and got the correct answer.
  • #1
Seydlitz
263
4

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:
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  • #2
Seydlitz said:

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
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
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
Ah thank you, I'll try so solve the problem and I'll let you know the result.
 

What is the Integration Method for Irrational Root?

The Integration Method for Irrational Root is a mathematical technique used to integrate functions containing irrational roots, such as square roots or cube roots.

Why is the Integration Method for Irrational Root important?

The Integration Method for Irrational Root is important because it allows us to solve complex integration problems that cannot be solved using basic integration techniques.

What are the steps involved in using the Integration Method for Irrational Root?

The steps involved in using the Integration Method for Irrational Root include identifying the irrational root, applying appropriate algebraic manipulations to the function, and using substitution or partial fractions to simplify the integral.

Can the Integration Method for Irrational Root be used for all types of irrational roots?

No, the Integration Method for Irrational Root is specifically designed for integrating functions with square roots or cube roots. Other types of irrational roots may require different integration techniques.

Are there any limitations to the Integration Method for Irrational Root?

Yes, the Integration Method for Irrational Root may not always provide an exact solution and may require approximations or numerical methods in some cases.

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