Why can I substitute x = a sin θ?

Question

I am reading my Calculus book, Calculus 8th Edition by James Stewart, and in 7.3 (pp. 526), it explains that I can use the reverse substitution: $$x=a\sin(\theta)$$ for the integral: $$\int{\sqrt{a^{2}+x^{2}}dx}$$ My question is, how does this work? Is it some special relationship or identity I am missing, or is it just an arbitrary substitution that can be made temporarily to calculate the integral? If the latter, does that mean I can substitute or reverse substitute anything when integrating as long as the integration is consistent, even if it's completely unrelated to the process or has no relationship to anything in the integrand? If not, or even if so, is there a relationship between: $$x \text{ and } a\sin(\theta)$$ If there is a relationship, can someone please explain it, or at least point me to a source that explains it well.

Answer 1

K.defaoite answered my question. It turns out I was looking at the concept the wrong way, and it's just an algebra trick for getting what you want in the integration process.

Answer 2

Are you sure that you read it careful? for $$\int{\sqrt{a^{2}-x^{2}}dx}\\x=a\sin(\theta) \to \int{\sqrt{a^{2}-a\sin^2(\theta)}a^2\cos(\theta)}d(\theta)=\\ \int a{\sqrt{1-\sin^2(\theta)}a\cos(\theta)}d(\theta)=$$and for $$\int{\sqrt{a^{2}+x^{2}}dx}\\x=a\tan(\theta)\to \int{\sqrt{a^{2}+a^2\tan(\theta)^{2}}d(1+\tan^2(\theta))d\theta}\\x=a\tan(\theta)\\ \int{a\sqrt{\underbrace{1+\tan(\theta)^{2}}_{\frac{1}{\cos^2(\theta)}}}d(1+\tan^2(\theta))d\theta}$$