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Wolfram Alpha's solution

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!

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    $\begingroup$ To get rid of the square root. Then use $\sin^2 x + \cos^2 x = 1$ .. $\endgroup$ – King Tut Mar 4 '18 at 12:26
  • $\begingroup$ what is your question? $\endgroup$ – Dr. Sonnhard Graubner Mar 4 '18 at 12:27
  • $\begingroup$ Trigonometric Substitution is a standard technique used in integration, we make those substitutions because often the derivatives of trigonometric functions are divisible by the functions rewritten through their pythagorean identities, allowing for some nice cancellation $\endgroup$ – Harry Alli Mar 4 '18 at 12:30
  • $\begingroup$ Dr Graubner click on the blue title which says "Wolfram Alpha's solution" I'm asking for insight into the approach used to solve it. To me the use of sin(u) and cos(u) seems totally random! I'd like to know why and how they were used. $\endgroup$ – James Mar 4 '18 at 12:32
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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.

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i think better is to write $$v^2\left(1-\left(\frac{kx}{v}\right)^2\right)$$ and to Substitute $$t=\frac{kx}{v}$$

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