This above photo is the working of an integration performed by WolframAlpha. I was hoping to gain some understanding rather than just an answer, in particular, why is it that it chose to substitute $x$ for $\dfrac{v\sin(u)}k $ and $dx$ for $\dfrac{v\cos(u)}k$?
I have a fairly good understanding of integration by substitution, is there a relationship between $\dfrac{1}{\text{sqrt}}$ and $\sin$ and $\cos$ waves I'm not familiar with? Hope you can help!
On
The substitution stems from the standard integral result for $$\int \dfrac{1}{\sqrt{a^2 - x^2}} \mathrm{d}x$$
Consider the simple case that $a = 1$, thus one can substitute $x = \sin\theta$ and proceed from there on using rudimentary trigonometric identities. Consider any arbitrary $a \in \mathbb{R}$, it should be easy for you to now derive the general result, i.e. $$\int \dfrac{1}{\sqrt{a^2 - x^2}} \mathrm{d}x = \arcsin \left(\dfrac{x}{a}\right) + C$$
Hope this helps. Cheers.
i think better is to write $$v^2\left(1-\left(\frac{kx}{v}\right)^2\right)$$ and to Substitute $$t=\frac{kx}{v}$$