I am given that $\mathbf{k} . m \mathbf{q} \wedge \mathbf{\ddot{q}} = 0$ and my book says that integrating this wrt time gives $\mathbf{k}.m \mathbf{q} \wedge \mathbf{\dot{q}} = $constant. I don't understand why this is. I've tried finding an explanation of integrating cross product but haven't really been able to find anything useful, this seems similar but I still can't figure out how it works.
Thank you.
So I assume that your $\lor$ is the cross product, which I usually write as $\times$.
If you have two vector functions $\vec{f}(t)$ and $\vec{g}(t)$ (with values in $\mathbb{R}^3$) , then you can form the cross product: $\vec{f}(t)\times \vec{g}(t)$. This is again a vector function. To take the derivative, the rule is that $$ \frac{d}{dt} \vec{f}(t)\times \vec{g}(t) = \frac{d}{dt}\vec{f}(t)\times \vec{g}(t) + \vec{f}(t)\times \frac{d}{dt} \vec{g}(t). $$ In other words it works just like the product rule for real valued functions.
Now, in your case you want to take the integral of a cross product. You can do this by verifying that the derivative of $\mathbf{k}.m \mathbf{q} \wedge \mathbf{\dot{q}}$ indeed is $\mathbf{k} . m \mathbf{q} \wedge \mathbf{\ddot{q}} = 0$.
First note that the $\bf{k}$ doesn't matter because it is a constant (see this). Likewise with the $m$. Now the other answer tells you exactly how you can do this.