Is there an easier way to integrate this equation using a triangle?

[Philosophaie]

I need to take the integral of an equation. I am trying to do Integration by Parts but I am coming up short.

I need to integrate (x^2 + y^2 + z^2)^(-3/2) dy for a Astronomy problem.
I tried:

u = x^2 + y^2 + z^2
du = 2*y dy

v=u^(-3/2)
dv = -3/2*u^(-5/2)

I can not do anything with the "y" in du.

Can you help?

 

[Mark44]

I need to take the integral of an equation.

First off, what you have below is NOT an equation -- an equation has two expressions with = in between.

I am trying to do Integration by Parts but I am coming up short.

I need to integrate (x^2 + y^2 + z^2)^(-3/2) dy for a Astronomy problem.

Like this?
$$\int (x^2 + y^2 + z^2)^{-3/2}dy$$

It would help to see the actual problem, to verify that the integral you set up is appropriate for the question. Assuming for the moment that you have the right integral, I don't think integration by parts is the way to go. I would use trig substitution instead.

I tried:

u = x^2 + y^2 + z^2
du = 2*y dy

v=u^(-3/2)
dv = -3/2*u^(-5/2)

I can not do anything with the "y" in du.

Can you help?

 

[HallsofIvy]

If x, y, and z are independent variables, as they normally are, this is just \int (a^2+ y^2)^{3/2} dy with a^2= x^2+ z^2. The substitution y= asin(\theta) shoud work.

 

[Mark44]

If x, y, and z are independent variables, as they normally are, this is just \int (a^2+ y^2)^{3/2} dy with a^2= x^2+ z^2. The substitution y= asin(\theta) shoud work.

The exponent is -3/2, per the OP.

 

[Philosophaie]

Would it not be y = a*tan(t)?
1
For

set

. In this case we talk about sine-substitution.
2
For

set

. In this case we talk about tangent-substitution.
3
For

set

. In this case we talk about secant-substitution.
a = \sqrt{x^{2} + z^{2}}

y = a*\tan{(t)}
dy=a*(/sec{(t)})^2 dt
t = \arctan{(\frac{x^{2} + y^{2} + z^{2}}{a})}

\int \frac{a*(\sec(t))^2}{(a^3 (\tan{(t)}^{3})}dt

I get lost from here.

 

[Mark44]

Would it not be y = a*tan(t)?

Yes, this would be the right trig substitution.

1
For

set

. In this case we talk about sine-substitution.
2
For

set

. In this case we talk about tangent-substitution.
3
For

set

. In this case we talk about secant-substitution.
a = \sqrt{x^{2} + z^{2}}

y = a*\tan{(t)}
dy=a*(/sec{(t)})^2 dt
t = \arctan{(\frac{x^{2} + y^{2} + z^{2}}{a})}

\int \frac{a*(\sec(t))^2}{(a^3 (\tan{(t)}^{3})}dt

I get lost from here.

The integral should be $\int \frac{a\sec^2(t)dt}{a^3 \sec^3(t)}$

If you draw a right triangle with t as the acute angle, and the opposite side is y, adjacent side is a, the hypotenuse is $\sqrt{y^2 + a^2}$. So $\sec(t) = \frac{\sqrt{y^2 + a^2}}{a}$. Solve this equation for $\sqrt{y^2 + a^2}$ that is in the denominator.

 

[Philosophaie]

\text{Where do you get "}(a*sec(t))^{3} \text{ in the denominator?}

The integral should be ∫asec 2 (t)dta 3 sec 3 (t) \int \frac{a\sec^2(t)dt}{a^3 \sec^3(t)}

 

[Mark44]

In the original integral, the denominator is $(y^2 + a^2)^{3/2}$, right? Based on the trig substitution you chose, and on the right triangle I described, $\sec(t) = \frac{\sqrt{y^2 + a^2}}{a} = \frac 1 a (y^2 + a^2)^{1/2}$, or $(y^2 + a^2)^{1/2} = a \sec(t)$. Use this to rewrite what you started with in the denominator.

 

[Philosophaie]

a*sec(t) \text{ does not equal }\sqrt{y^2 + a^2}
a*\tan{(t)} = \sqrt{y^2 + a^2}

 

[Mark44]

a*sec(t) \text{ does not equal }\sqrt{y^2 + a^2}
a*\tan{(t)} = \sqrt{y^2 + a^2}

No, this is not how things are in my triangle, which I described in an earlier post.

In my triangle, tan(t) = y/a and $\sec(t) = \frac{\sqrt{y^2 + a^2}} a$

Have you actually drawn this triangle? I never bothered memorizing the formulas you wrote in post #5 -- I just draw a right triangle and label the sides and hypotenuse according to whether the expression in the radical is a sum or difference. If you don't use the triangle, you're doing things the hard way, IMO.