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chexmix
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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:
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]
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).
Thanks,
Glenn
Homework 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]
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).
Thanks,
Glenn
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