How come when using trig substitution you don't get modulus when squaring a square root?

Question

anonymous · Answer

Could you provide a example, please?

anonymous · Answer

well, you might have 1/sqrt(x^2+9) and you could substitute x=3tan(u) so when adding up the result inside the sqrt you get 9 tan^2(u)+9 which is 9(tan^2(u)+1) = 9 sec^2(u).. When extracting the squareroot from 9 sec^2(u) they get 3 sec(u)

anonymous · Answer

why not 3* |sec(u)|

anonymous · Answer

I mean that when using trig substitution they assume that
$$\sqrt{\tan ^{2}(x)}=\tan(x)$$ rather than$$|\tan(x)|$$

anonymous · Answer

I think that generally you have a point.  However, trig. sub. can go very wrong if your limits of integration would cause a domain/range problem with the trig function that you are using.  Both of the trig functions that are involved here have undefined points within their domain.  An integral evaluated over that interval would be invalid.  If you have limits of integration, you should check that all is well with the trig function that you are using over the given domain.  If you are evaluating a definite integral, you assume that you are operating in the first quadrant so that you avoid any such problems.  tan(x) and sec(x) are positive and defined in the first quadrant.

anonymous · Answer

I know some instructors who don't even mention this, but my instructor considered it worth one point on a test question to mention "assume first quadrant" so that you can remove the absolute value signs.

anonymous · Answer

I was looking at this problem
$$f (t)=\sqrt{1-(\sin 2t)^{2} }$$,$$-\pi/4<t<\pi/4$$
where the derivative was asked,
$$\sqrt{1-(\sin 2t)^{2}}=\cos 2t$$
however there is an issue about whether this is + or - $$  \cos 2t$$
The condition that t is between $$-\pi/4$$ and $$\pi/4$$
insures it  is a $$+\cos2t$$,the graph is actually positive,
so sin graph will be  between $$0<t<\pi$$,
tan $$0<t<\pi/2$$
basically it depends on the  condition if it is not explicit then the modulus is proper to apply here.

anonymous · Answer

So i guess that the necessary condition of assuming first quadrant is left of by many teachers for the purpose of what??? Not confusing the students??? Anyway, thanks guys what you're saying makes sense.