# Integrating factor to solve this?

1. Mar 15, 2017

### binbagsss

1. The problem statement, all variables and given/known data
$cos t \frac{dv}{dt} + (sin t) t = \frac{GM}{b^2 }\sin^3 t$

2. Relevant equations

above

3. The attempt at a solution

im pretty stuck to be honest. It almost looks like a product rule on the LHS but it has the wrong sign, RHS I've tried writing $sin^3 t$ as $(1-cos^2t)\sin t$ etc, pretty unsure what integration factor I need.

2. Mar 15, 2017

### Buzz Bloom

Hi binaqsss:

I suggest trying the substitution x = sin(t).

Regards,
Buzz

3. Mar 15, 2017

### Ray Vickson

If you wrote the problem correctly, then you just have
$$\frac{dv}{dt} = - t \tan t + a \frac{ \sin^3 t }{\cos t} = - t \tan t + a \sin^2 t \, \tan t,$$
so $v = - \int t \tan t \, dt + a \int sin^2 t \, \tan t \, dt$. The integration in the first term involves some non-elementary functions.

4. Mar 15, 2017

### LCKurtz

Did you perhaps mistype it and mean instead:$$\cos t \frac{dv}{dt} + (\sin t) \color{red}{v} = \frac{GM}{b^2 }\sin^3t \text{ ?}$$