I have this problem, a projectile is fired into fluid with a rate of deceleration $a=-0.4v^3$ and in initial velocity $v_0=60$.
We are to find how fast its going after $t=4$ seconds. If one starts with: $a=\frac{dv}{dt}=-.4v^3$, and advance to $dt=\frac{dv}{a}$, we can integrate $$\int_0^tdt=\int_{60}^v\frac{1}{-0.4v^3}$$
Integrating, this produces a correct answer of: $v=0.559$. However, when I set up an even simpler (looking) equation, the problem becomes: $adt=dv$, such that $-0.4v^3dt=dv$. The integral becomes: $$\int_0^t-0.4v^3dt=\int_{60}^vdv$$ solving for this equation produces a different, incorrect answer. What causes this?
In the second equation you propose you are integrating v^3 in dt on the l.h.s. but you cannot do that since v is an unknown function of the time. You can't just assume it doesn't depend on t. The "correct" expression of what you wrote should be $$\int_0^t -0.4v^3(t)dt=\int_{60}^vdv\ ,$$ but that's simply an involved way to write the differential equation $$\frac{dv(t)}{dt}=-0.4v^3(t)$$ with initial condition $v(0)=60$.