A simple antidifferentiation question

[chexmix]

I am brushing up my single-variable calculus, partly by working my way through the 9th edition of Thomas and Finney's Calculus and Analytic Geometry. I'm finding myself stuck at an early problem in antidifferentiation:

Homework Equations

a) \int sec^{2}x dx = tan x + C

b) \int \frac{2}{3} sec^{2} \frac{x}{3} dx = 2 tan (\frac{x}{3}) + C

The Attempt at a Solution

The first of these (a) makes sense since it was established earlier in the book that the derivative of tan x is sec^2 x. However, getting from problem to solution in (b) is confounding me and I am sure I am missing something very simple.

I tried researching this with Wolfram Alpha, and the steps it used to reach the solution included integration by substitution, a topic that has not been covered yet in Thomas / Finney.

Is there a simpler way to antidifferentiate (b)? My first step is to move the constant in front:

\frac{2}{3} \int sec^{2} \frac{x}{3} dx

... but after that I don't see a way besides substitution (which I remember from my first pass through this material over a year ago).

Thanks,

Glenn

 

[Screwdriver]

Well my good man, if you can't make a substitution, you can one of its crude forms, which is basically just guessing and checking. You know that \int sec^{2}x dx = tan x + C, so it stands to reason that \int sec^{2}(\frac{x}{3}) dx = tan (\frac{x}{3}) + C doesn't it? This isn't actually true though, so what you have to do now is differentiate tan (\frac{x}{3}) + C and see what you need to multiply it by to "fix" it. Remember that you can always check anti-differentiation by differentiation, so try that!

 

[chexmix]

Screwdriver,

Thanks for the reply. I guess I was looking for some bit of magic that doesn't exist!

 

[Screwdriver]

You're welcome, chexmix!

I would highly suggest learning substitutions though. They're pretty easy to understand and it removes the guessing aspect

 

[tastybrownies]

Yeah I would DEFINITELY learn u substitutions. I'm about 3/4 through the way of my Calc II class and I couldn't imagine not being able to use these u subs.

 

[chexmix]

Substitutions are definitely on my list.

I have had Calc I and II (though it was a year ago) and am now reviewing everything for a stab at Calc III in the Fall. I'm running out of review time, so things are getting a little frantic!

 

[tastybrownies]

I'm just trying everything I can right now to get through Calc II, that class is seriously a nightmare.

 

[chexmix]

I can't say I made a fantastic showing in either Calc I or II.

... but I had been away from math for 28 years when I started again with Pre-Calc a couple of years ago, so I try to be kind to myself.