I was trying to solve the following question
A particle moves in a straight line experiencing an acceleration equal to $\frac{2}{x^2}$ towards the origin. If the particle starts from rest at a distance $4$ units from the origin, show that the time it takes to reach the origin is $2\pi$
Context I placed the particle on the left of the origin, and said that it was moving towards the origin from there i.e. direction towards origin is positive. I said how its direction of motion is positive, and thus its acceleration is also positive (in this scenario). I proceeded from there as follows;
$$\ddot x=\frac{2}{x^2}=v\frac{dv}{dx} $$
As initially ($t=0$) $v=0$ (at rest) $x=-4$ (distance from origin), I evaluate the following; $$\int_0^v vdv = 2\int_{-4}^x \frac{dx}{x^2}$$ Solving, and rearranging $$v^2 = \frac{-x-4}{x}$$ $$\therefore v=+\sqrt{\frac{-x-4}{x}}$$ (As the velocity is towards the origin, which I said was positve, velocity is then positive).
As $v=\frac{dx}{dt}$, I rearrange the obtain the following integral (Note: at $x=-4, t=0$ and $x=0, t=T_1$ (desired time)); $$\int_{-4}^0 \frac{\sqrt{x}}{\sqrt{-x-4}}dx = \int_0^{T_1} dt$$
Though I currently don't know of any method to easily solve the integral on the right hand side. Wolfram alpha evaluates it as $2\pi$, but regardless I don't know how to easily solve the integral.
BUT
If I defined the motion of the particle as negative (then acceleration is negative) i.e. direction away from origin positive, I eventually obtain the following integral; $$\int_{4}^0 \frac{\sqrt{x}}{\sqrt{4-x}}dx = - \int_0^{T_1} dt$$
Which can easily be solved through a simple trigonometric substitution.
The Question Why does changing the direction which is considered positive, DRAMATICALLY effect the difficulty of the integral to be evaluated? For motion questions (or maybe just in general), how would I know beforehand which direction to make positive, so as to eliminate the need to evaluate a more complicated integral in trade for a much more simplistic one (as above)?
Thanks
Look again at the integral $$\int_{-4}^0 \frac{\sqrt{x}}{\sqrt{-x-4}}dx.$$
Do you see what's wrong with it? You're integrating over $x \in [-4,0],$ but the integrand isn't defined anywhere in that interval: $-x-4 < 0$ when $x \leq 0,$ and you can't take the square root of a negative number (at least not in real analysis).
Let's try it again. Presumably, you got that integral by substituting $v=\sqrt{\frac{-x-4}{x}}$ in the integral $\int_{-4}^0 \frac{dx}{v}.$ That is, $$\int_{-4}^0 \frac{dx}{v} = \int_{-4}^0 \sqrt{\frac{x}{-x-4}}\,dx.$$
This is fine, because dividing a negative by a negative gives you a positive number, and you can take the square root of a positive number. In order to rewrite $\sqrt{\frac ab}$ as $\frac{\sqrt a}{\sqrt b},$ however, $a$ must be non-negative and $b$ must be positive. So let's make them so: $$ \sqrt{\frac{x}{-x-4}} = \sqrt{\frac{-x}{x+4}} = \frac{\sqrt{-x}}{\sqrt{4+x}}.$$ Note that for $-4 < x < 0,$ both $-x$ and $4+x$ are positive, so these equations are correct.
So now you can write $$ \int_{-4}^0 \sqrt{\frac{x}{-x-4}}\,dx = \int_{-4}^0 \frac{\sqrt{-x}}{\sqrt{4+x}}\,dx = -\int_4^0 \frac{\sqrt{u}}{\sqrt{4-u}}\,du $$ with the substitution $u = -x,$ and you have your "easy" integral.