# Integration problem?

kashiark

∫ 1/r³ dr/dt

## Homework Equations

∫ xa dx = x(a+1)/(a+1)

## The Attempt at a Solution

I have no clue how to solve it like this. I don't have the equation for r in terms of t, so I can't just substitute. How would I do it?

Homework Helper

∫ 1/r³ dr/dt

## Homework Equations

∫ xa dx = x(a+1)/(a+1)

## The Attempt at a Solution

I have no clue how to solve it like this. I don't have the equation for r in terms of t, so I can't just substitute. How would I do it?

is that how the question is actually written & what are you integrating with respect to?

it doesn't really make sense how you have written it, do you mean:
$$\int dt (\frac{1}{r(t)^3}\frac{dr}{dt})$$

if so have a think about how chain rule differentiation works & how you could try & reverse it using FTC...

kashiark
OK if it were like that, wouldn't the dt's just divide out allowing you to integrate with respect to r?
∫ dt(1/r³ dr/dt)
∫ 1/r³ dr
-1/(2r²) + C

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Homework Helper
you can, but its a bit of an abuse of notation, athough it gives the same answer, a better way to think of it is to write:

$$\frac{d}{dt} (-\frac{1}{2r(t)^2}) = \frac{1}{r(t)^3}\frac{dr}{dt}$$

then the intergal becomes
$$\int dt (\frac{d}{dt} (-\frac{1}{2r(t)^2}))$$

so by FTC, the anti-derivative is
$$= (-\frac{1}{2r(t)^2}) +C$$

though the question was to you - how is it written in the actual problem?

kashiark
It's not on a worksheet or anything; I was just messing with a particular physical situation, and I came to something in that form that I needed to integrate.