The discussion revolves around understanding a specific integral involving trigonometric substitution, particularly the integral ∫[x^2*dx/sqrt(9+x^2)]. Participants are examining the implications of substituting x = 3tan(t) and the subsequent derivative transformations.
The discussion is active, with participants questioning the clarity of explanations and the choice of notation. There is a mix of perspectives on the use of trigonometric functions and their derivatives, indicating a productive exploration of different approaches to the problem.
Some participants express familiarity with different notational conventions for trigonometric functions, which may influence their understanding of the derivative explanations provided. There is an underlying assumption that participants have a foundational knowledge of calculus and trigonometry.
If ##x = 3\tan(t)##, then ##dx = 3\sec^2(t)dt = 3(\tan^2(t) + 1)dt##Memo said:
That was just from the step of taking the derivative of sin/cos.Mark44 said:
I guess aka ##sec^2 tdt##FactChecker said:
Why not just go directly to the derivative of the tangent function?FactChecker said:
I don't know. Your guess is as good as mine. Maybe they wanted to keep the required background to just derivative of sin, cos, and the quotient rule.Mark44 said:
The derivative of ##tan(x)## is ##\frac1{cos^2(x)}=sec^2(x)##.Mark44 said:
Yes, I realize that. My point was in explaining why ##x = 3\tan(t)## implies that ##dx = 3\sec^2(t)dt = 3(1 + \tan^2(t))dt##.martinbn said:
What was problem with the explanation ##dx = \frac{3dt}{cos^2(t)} = 3(1 + \tan^2(t))dt##.Mark44 said:
Assuming someone knows the derivatives of all six circular trig functions, why would someone write ##\frac{d\tan(t)}{dt}## as ##\frac 1 {\cos^2(t)}## and not just go directly to ##\sec^2(t)##?martinbn said:
Where I come from ##sec## and ##csc## are not used at all. For someone like me the derivative of tangent is one over cosine squared.Mark44 said:
I have taught college-level mathematics for 20+ years. Every calculus textbook I've ever seen presents all six circular trig functions as well as their derivatives.martinbn said:
Yes, and when I went to the US and thought calculus, I learned the notations for ##\frac1{\cos(x)}## and ##\frac1{\sin(x)}##. But where I grew up there were no such functions.Mark44 said: