[Prefix] When we do trigonometric substitutions (such as for the integral x^3/(a^2-x^2)^2), we say something like "let x = asinp for -pi/2 <= p <= pi/2" then we carry on and solve the integral. However, sometimes our answer is ugly and we get some term in our expression like "cosp"- so we draw a little right angle triangle (SOH CAH TOA) to find out that "cosp = sqrt(a^2-x^2)/a <==> p = arccos(sqrt(a^2-x^2))" We then substitute this into cosp to get cos(arccos(Q(x))) = Q(X). Now our answer is nice and it is expressed in terms of x again. [My Question] Do we have to place a domain restriction on our answer when we do this? Because to me it is pretty clear that the SOH-CAH-TOA triangle only gives us valid results for x>0 and p>0 What I mean is that sinp = x/a and cosp = sqrt(a^2-x^2)/a is only true for p>=0 But beforehand we already defined our domain to be -pi/2 <= p <= pi/2 (when we did our sine substitution). However, we got "cosp" in our answer, so we have to express it in terms of x using arccos (but as stated above, this is only true for p>0) So is it true that if we use "cosp = sqrt(a^2-x^2)" anywhere we now have to make our new domain p <= pi/2 ? In all of the trig substitution questions I've done in calculus related courses, they've never asked for the domain, it was more of a mindless drone thing to do- and I'm really curious about this question.