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I originally posted this as a second part to a messy slew of questions I had on some lecture notes I had, so I decided to take the suggestion of some users and post this question individually.

I have a question I find I can't properly explain. In any integration by trig substitution problem you have to state something along the lines of this, for example:

$x = sin\ u$

What is this insinuating? Why am I allowed to even do this without it changing the problem entirely in the same way that this substitution is invalid?:

  • $x + 4 = 10$

  • Let $x = sin\ y$

  • Therefore $sin\ y = 6$

Why is the above not okay other than the obvious that $sin\ y = 6$ is impossible yet it's okay in integration by trig substitution? Is that because of our changing the differential variable?

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In cases where the substitution would require you to be able to say that sometimes $\sin y = 6,$ the substitution is not OK.

For example, you wouldn't make the substitution $x = \sin y$ in order to integrate $\int x^2 \,dx,$ partly because you already know how to do that integral without the substitution, but also because $x = \sin y$ would give you an integral that is valid only for $-1\leq x \leq 1,$ whereas the original integral is valid for all $x.$

On the other hand, if you have to integrate $$ \int \sqrt{1 - x^2} \,dx,$$ then the integral is already valid only for $-1\leq x \leq 1,$ because for $\lvert x\rvert > 1$ we would have $1 - x^2 < 0,$ and the square root of a negative number takes us outside the domain of real-number integration. So in that case you don't lose anything by substituting $x = \sin y.$

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  • $\begingroup$ I see, so by making the $sin\ y$ substitution I'm implementing a domain restriction, since −1≤$sin\ y$≤1, and this would not accurately represent the integral at values other less than -1 or greater than 1. But since, like in your example, the function's domain already only lies there, such a substitution could accurately still represent the function? Hopefully this question makes sense. Like, essentially −1≤x≤1 and −1≤$sin\ y$≤1 so $sin\ y = x$? $\endgroup$ Commented Jun 12, 2017 at 17:31
  • $\begingroup$ Yes, that is the idea. It works as you said. $\endgroup$ Commented Jun 12, 2017 at 18:46

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