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A simple antidifferentiation question

  1. Aug 16, 2011 #1
    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:

    2. Relevant equations

    a) [itex]\int sec^{2}x dx = tan x + C[/itex]

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

    3. The attempt at a solution

    The first of these (a) makes sense since it was established earlier in the book that the derivative of [itex]tan x[/itex] is [itex]sec^2 x[/itex]. 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:

    [itex]\frac{2}{3} \int sec^{2} \frac{x}{3} dx[/itex]

    ... 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).


    Last edited: Aug 16, 2011
  2. jcsd
  3. Aug 16, 2011 #2
    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 [itex]\int sec^{2}x dx = tan x + C[/itex], so it stands to reason that [itex]\int sec^{2}(\frac{x}{3}) dx = tan (\frac{x}{3}) + C[/itex] doesn't it? This isn't actually true though, so what you have to do now is differentiate [itex]tan (\frac{x}{3}) + C[/itex] 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!
  4. Aug 16, 2011 #3

    Thanks for the reply. I guess I was looking for some bit of magic that doesn't exist! :biggrin:
  5. Aug 16, 2011 #4
    You're welcome, chexmix!

    I would highly suggest learning substitutions though. They're pretty easy to understand and it removes the guessing aspect :smile:
  6. Aug 16, 2011 #5
    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.
  7. Aug 17, 2011 #6
    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! :eek:
  8. Aug 18, 2011 #7
    I'm just trying everything I can right now to get through Calc II, that class is seriously a nightmare.:yuck:
  9. Aug 18, 2011 #8
    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.
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