Solving Integrals with Square Root in Denominator

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Homework Help Overview

The discussion revolves around solving an integral involving a square root in the denominator, specifically the integral of the form \(\int_{0}^{a}\frac{s \ ds}{\sqrt{y^{2} + s^{2}}}\). Participants are seeking clarification on the steps taken to evaluate this integral and the reasoning behind the use of substitutions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the potential use of substitution in solving the integral, with one suggesting \(u = y^2 + s^2\) and questioning how to determine when such substitutions are appropriate. Others express confusion about the transition from the integral to the evaluated expression.

Discussion Status

The discussion is ongoing, with participants sharing their attempts and seeking further clarification on the substitution method. There is an acknowledgment of the need for practice in recognizing when to apply specific techniques, but no consensus has been reached on the best approach.

Contextual Notes

Participants are grappling with the concept of substitution in integrals and the conditions under which it is applicable. There is a mention of the importance of practice in developing intuition for these types of problems.

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Homework Statement



I need to proceed to solve this integral:

Homework Equations



[tex]\int_{0}^{a}\frac{s \ ds}{\sqrt{y^{2} + s^{2}}} = \left [ \sqrt{y^{2} + s^{2}} \right ]_{s=0}^{s=a}[/tex]

The Attempt at a Solution



I don't understand what he did from the left-hand side of the equation to the right-hand side... could someone explain me, please?

Thank you,
Rafael Andreatta
 
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Taturana said:

Homework Statement



I need to proceed to solve this integral:

Homework Equations



[tex]\int_{0}^{a}\frac{s \ ds}{\sqrt{y^{2} + s^{2}}} = \left [ \sqrt{y^{2} + s^{2}} \right ]_{s=0}^{s=a}[/tex]

The Attempt at a Solution



I don't understand what he did from the left-hand side of the equation to the right-hand side... could someone explain me, please?

Thank you,
Rafael Andreatta
It seems to me that an ordinary substitution was used: u = y2 + s2, du = 2s ds.
 
Mark44 said:
It seems to me that an ordinary substitution was used: u = y2 + s2, du = 2s ds.

I tried some math here and I still don't understand...

When you get an integral of that type, how do you know when you need to make this kind of substitution?

Could you explain me more clearly?

Thank you very much...
 
Taturana said:
I tried some math here and I still don't understand...

u=y2+s2

if y is a constant and you differentiate wrt s you will get

du/ds = 2s or du=2s ds, meaning that s ds = du/2

Taturana said:
When you get an integral of that type, how do you know when you need to make this kind of substitution?

It comes with practice really. If you have this integral

[tex]\int \frac{x^2}{x^3+1} dx[/tex]

you notice that d/dx(x3+1) = 3x2

so if you used a substitution of t=x3+1, dt = 3x2 dx or dt/3 = x2 dx.

So in the integral, you can replace x2 dx with dt/3.
 
Taturana said:
I tried some math here and I still don't understand...

When you get an integral of that type, how do you know when you need to make this kind of substitution?

Could you explain me more clearly?

Thank you very much...
Show us what you're trying and we'll straighten you out.
 

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