please delete

[lisasayzhi]

please delete

 

[royhaas]

I didn't check them closely, but your approach is correct.

 

[soroban]

Re: Trigonometric substitutions

Hello, lisasayzhi!

I couldn't follow your work all the through, sorry.
$\displaystyle \;\;$But your work is truly impressive!
You could have save a LOT work with one observation . . .

\(\displaystyle \text{1. Find }\L I\;=\;\int \frac{1}{\cos^3x}\,dx\)

$\displaystyle u\,=\,\frac{1}{\cos x}\;\;\Rightarrow\;\;du\,=\,\frac{\sin x}{\cos^2x}\,dx$

$\displaystyle dv\,=\,\frac{1}{\cos^2x}\,dx\;\;\Rightarrow\;\;v\,=\,\frac{\sin x}{\cos x}$

\(\displaystyle \L I\;=\;\frac{\sin x}{\cos^2x} - \int \frac{\sin^2x}{\cos^3x}dx\)

\(\displaystyle \L I\;=\;\frac{\sin x}{\cos^2x} - \int \frac{1\,-\,\cos^2x}{\cos^3x}dx\)

\(\displaystyle \L I\;=\;\frac{\sin x}{\cos^2x} - \underbrace{\int \frac{1}{\cos^3x}dx} + \int \frac{1}{\cos x}dx\;\;\) . . . Here!
. . . . . . . . . . . . . . . this is I !

So we have: \(\displaystyle \L\,I\;=\;\frac{\sin x}{\cos^2x}\,-\,I\,+\,\int\frac{1}{cos x}dx\)

Then: \(\displaystyle \L\,2I\;=\;\frac{sin x}{\cos^2x}\,+\,\int\frac{1}{\cos x}dx\;=\;\frac{\sin x}{\cos^2x}\,+\,\int \sec x\,dx\)

\(\displaystyle \L\;\;\;2I\;=\;\frac{sin x}{\cos^2x}\,+\,\ln|\sec x\,+\,\tan x| + C\)

\(\displaystyle \L\;\;\;I\;=\;\frac{1}{2}\left[\frac{\sin x}{\cos^2x}\,+\,\ln|\sec x\,+\,\tan x|\right]\,+\,C\)

Therefore: \(\displaystyle \L\,\int\frac{dx}{\cos^3x}\)$\displaystyle \;=\; \frac{1}{2}[\sec x\tan x \,+\,\ln|\sec x\,+\,\tan x|]\,+\,C$

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By the way, you just integrated a Classic problem: $\displaystyle \,\int \sec^3x\,dx$