I am trying to understand projectile motion, but I am stuck. I'm trying to follow http://farside.ph.utexas.edu/teaching/336k/Newtonhtml/node29.html#e5.23u. However, I don't understand how they reach equation 178 by integrating equation 176. And similarly how they reach equation 181 by integrating equation 177.
EDIT: I've realised vt is a constant. Therefore you have integral of (1/vx)dt between vx0, v which = (-g/vt) * integral (1/dt).
So ln(v/vx0) = (-g/vt)*t + c.
So where does the c go?
Starting from equation 176:
$$ \frac{dv_x}{dt} = - g \frac{v_x}{v_t},$$
we move around the factors
$$ \frac{1}{v_x} \frac{dv_x}{dt } = - \frac{g}{v_t},$$
and we integrate both sides to $dt$, with boundaries $t = 0$ and $t = T$.
$$ \int_0^T \frac{1}{v_x} dt \frac{dv_x}{dt} = - g \int_0^T \frac{dt}{v_t}.$$
We can use the chain rule to change the integration variable in the left integral. This also changes the boundaries:
$$ \int_{v_x \mathrm{\ when\ } t = 0}^{v_x \mathrm{\ when\ } t = T} \frac{d v_x}{v_x} = - g \frac{(T - 0)}{v_t}.$$
Calculating the integral leaves us with
$$ \ln(v_x(T)) - \ln(v_x(0)) = -g \frac{T}{v_t}.$$
The $c$ is hidden in the $v_t$ (note its definition).
Also remark that your source has the horrible habit of denoting both variables and constants by the same letter ($v_x, t$), probably being the source for most of your confusion.