Question

∫ sec(x) dx=?
Please just guide me along, because I want to get that sense of accomplishment of figuring something out.
∫ sec(x) dx -> ∫ (sec(x)tan(x))/(tan(x) dx
u=sec x
du/dx=sec x tan x
dx = du/(sec x tan x)
--> ∫u/u' du
Then, integration by parts
f(x)=u, g'(x)=1/u'
u ln u' - ∫ 1*ln u' du
=u ln u' -u' ln u' - u'
=(u-u')ln u' -u'
resubstitute
(sec x-sec x tan x)ln(sec x tan x)- sec x tan x
I know I forgot the abs signs some where, and I'm wondering if I even did this right.

Answer 1

Also, thanks a lot Wolfram Alpha, for not explaining this at all. -_-

Answer 2

multiply by a useful form of "1"

sec+tan
------
sec+tan

Answer 3

one issue i see that might be throwing you for a loop is that sec tan is not defined in us

Answer 4

Int Secx => you have to multiply Secx by \[(\sec x+\tan x)/(\sec x+\tan x)\] in order to have the derivative of the bottom on top. You should get:
\[\int\limits_{?}^{?}(\sec ^{2}x+\sec x \tan x)/(\sec x+\tan x)\]
Now you can just do u-sub u=secx + tanx du = (sec x tan x +Sec^2)dx
You will end up with:\[\int\limits_{?}^{?}\sec x dx=\ln |\sec x+\tan x|+C\]

Answer 5

you havent learned to type in latex yet i see

Answer 6

Ha no, too complicated

Answer 7

It was sec x+tan x?! FUUUUUUUUUUUUUUUUU

Answer 8

\int for \(\int\)
\frac{n}{d} for \(\cfrac{n}{d}\)

Answer 9

Wait, is my integration correct though? I know it was a ridiculously convoluted way, but...

Answer 10

i dont think so; you introduced a flaw in it at the beginning i think that persisted towards the end

Answer 11

dx = du/sec(x) tan(x) would be fine ... but youd have to express tan(x) in terms of u as well