# Integral of an equation

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1. Sep 4, 2015

### 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?

2. Sep 4, 2015

### Staff: Mentor

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

3. Sep 4, 2015

### HallsofIvy

Staff Emeritus
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.

4. Sep 4, 2015

### Staff: Mentor

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

5. Sep 4, 2015

### 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.

6. Sep 5, 2015

### Staff: Mentor

Yes, this would be the right trig substitution.

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.

7. Sep 5, 2015

### Philosophaie

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

Last edited: Sep 5, 2015
8. Sep 5, 2015

### Staff: Mentor

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.

9. Sep 5, 2015

### Philosophaie

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

Last edited: Sep 5, 2015
10. Sep 5, 2015

### Staff: Mentor

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.