Solving Integrals with Square Root in Denominator

[Taturana]

Homework Statement

I need to proceed to solve this integral:

Homework Equations

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

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

 

[Mark44]

Homework Statement

I need to proceed to solve this integral:

Homework Equations

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

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.

 

[Taturana]

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...

 

[rock.freak667]

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

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

\int \frac{x^2}{x^3+1} dx

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.

 

[Mark44]

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.