Question

anti derivative of sec(x)?

Answer 1

@zepdrix ? :)

Answer 2

So um,
this requires a very non intuitive trick.
It's not something that you would come up with on your own,
so I can understand why you're stuck lol.

Answer 3

\[\large\rm \int\limits \sec x~dx\]Multiply by a fancy version of 1,
(secx+tanx)/(secx+tanx)

Answer 4

\[\large\rm \int\limits \sec x\frac{\sec x+\tan x}{\sec x +\tan x}dx\]

Answer 5

\[\large\rm \int\limits \frac{\sec x \tan x +\sec^2x}{\sec x+\tan x}dx\]Then it's a simple u-substitution from there :)
u = denominator.

Answer 6

Then it's just
(1/u)du

Answer 7

Good good good.

Answer 8

ln|u|+c
ln|sec(x)+tan(x)|+C

Answer 9

yay \c:/

Answer 10

yay!! thanks :D that was a bit of a stretch..multiplying it by that fancy 1. I would never have thought of it o.0

Answer 11

There is an "intuitive" way, but it is longer.

Answer 12

\(\color{#000000}{ \displaystyle \int \frac{1}{\cos x}dx =\int \frac{\cos x}{1-\sin^2 x}dx=\int \frac{1}{1-z^2}dz =\frac{1}{2}\int \left(\frac{1}{z+1}-\frac{1}{z-1}\right)dz }\)

\(\color{#000000}{ \displaystyle =\frac{1}{2} \ln\left|z+1\right|-\frac{1}{2}\ln\left|z-1\right|+C}\)

\(\color{#000000}{ \displaystyle =\frac{1}{2} \ln\left|\sin x+1\right|-\frac{1}{2}\ln\left|\sin x-1\right|+C}\)

\(\color{#000000}{ \displaystyle =\frac{1}{2} \ln\left|\frac{\sin x+1}{\sin x-1}\right|+C =\frac{1}{2} \ln\left|1+\frac{2}{\sin x-1}\right|+C}\)

Wait, did I just now get a different answer?

Answer 13

mm yeah that's quite different xD

Answer 14

Either it's wrong, but I don't see a mistake, or I will have to apply my skills in trig, if I got any:)

\(\color{#000000}{ \displaystyle \sqrt{\frac{\sin x+1}{\sin x-1}}=\sec x+\tan x}\)

Answer 15

\(\color{#000000}{ \displaystyle \sqrt{\frac{1+\sin x}{1-\sin x}}=\frac{1+\sin x}{\cos x} }\)

\(\color{#000000}{ \displaystyle \sqrt{\frac{1+\sin x}{1-\sin x}}=\frac{1+\sin x}{\sqrt{1-\sin^2x}} }\)

\(\color{#000000}{ \displaystyle \sqrt{\frac{1+\sin x}{1-\sin x}}=\frac{1+\sin x}{\sqrt{1-\sin x}\sqrt{1+\sin x}} }\)

\(\color{#000000}{ \displaystyle \sqrt{\frac{1+\sin x}{1-\sin x}}=\frac{\sqrt{1+\sin x}}{\sqrt{1-\sin x}} }\)

\(\color{#000000}{ \displaystyle \sqrt{\frac{1+\sin x}{1-\sin x}}=\sqrt{\frac{1+\sin x}{1-\sin x}} }\)

Answer 16

Yeah, they are the same thing actually.

Answer 17

No, don't worry about this, just go with what zep said.

Answer 18

Thanks:) your way is cool too! xP