Help understanding this integral solution using trig substitution please

[Memo]

Homework Statement

∫[x^2*dx/sqrt(9+x^2)]

Relevant Equations

sqrt(1+tan^2(a))~sqrt(a^2+U^2)

Here's the answer:

Could you explain the highlighted part for me? Thank you very much!

 

[Mark44]

Homework Statement: ∫[x^2*dx/sqrt(9+x^2)]
Relevant Equations: sqrt(1+tan^2(a))~sqrt(a^2+U^2)

View attachment 334648
Here's the answer:
View attachment 334650
Could you explain the highlighted part for me? Thank you very much!

If $x = 3\tan(t)$, then $dx = 3\sec^2(t)dt = 3(\tan^2(t) + 1)dt$
The part where they have $\frac{3dt}{\cos^2(t)}$ isn't very helpful, IMO. Better to just replace $\sec^2(t)$ with $1 + \tan^2(t)$.

 

[FactChecker]

The part where they have $\frac{3dt}{\cos^2(t)}$ isn't very helpful

That was just from the step of taking the derivative of sin/cos.

 

[WWGD]

That was just from the step of taking the derivative of sin/cos.

I guess aka $sec^2 tdt$

 

[Mark44]

That was just from the step of taking the derivative of sin/cos.

Why not just go directly to the derivative of the tangent function?

 

[FactChecker]

Why not just go directly to the derivative of the tangent function?

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.

 

[martinbn]

Why not just go directly to the derivative of the tangent function?

The derivative of $tan(x)$ is $\frac1{cos^2(x)}=sec^2(x)$.

 

[Mark44]

The derivative of $tan(x)$ is $\frac1{cos^2(x)}=sec^2(x)$.

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]

Yes, I realize that. My point was in explaining why $x = 3\tan(t)$ implies that $dx = 3\sec^2(t)dt = 3(1 + \tan^2(t))dt$.

What was problem with the explanation $dx = \frac{3dt}{cos^2(t)} = 3(1 + \tan^2(t))dt$.

 

[Mark44]

What was problem with the explanation $dx = \frac{3dt}{cos^2(t)} = 3(1 + \tan^2(t))dt$.

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 $\sec^2(t)$?
I'm not saying what was written was incorrect, just that it was more complicated than was necessary.

 

[martinbn]

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 $\sec^2(t)$?
I'm not saying what was written was incorrect, just that it was more complicated than was necessary.

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]

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.

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]

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.

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.