Question

How do I get the integral for sec x.

Answer 1

\[\int \sec x \; dx \implies \int \sec x \times \frac{\sec x + \tan x}{\sec x + \tan x} \; dx \implies \in \frac{\sec^2 x + \sec x \tan x}{\sec x + \tan x} dx\]

\[\implies \ln \; | \; \sec x + \tan x \; | \]
does that help?

Answer 2

uhh that should be \(\int\) not \(\in\) lol

Answer 3

obviously,giving directly the solution will help...

Answer 4

@lgbasallote read this:
http://openstudy.com/code-of-conduct

Answer 5

Iventer is asking *how* to get the integral...obviously you show *how* won't you?

Answer 6

i have heard that here people make the asker reach the solution....but one very important member is not doing so,i guess.....

Answer 7

@lgbasallote tell why did u multiply and divide by sec x + tan x.....

Answer 8

so do you believe that beating around the bush and giving indirect answers is more effective than showing how?

Answer 9

it will just confuse people more

Answer 10

Yes, thank you Igbasallote. I had the answer to the integral (was given by textbook), but just did not know how to get to it, and I wanted to know how they got it. Much appreceated xxx

Answer 11

ok,so this is answering site,not learning site.....
hmmmm.....i m out of it....

Answer 12

And i know now why u multiplied with "1" :sec + tan

Answer 13

@gokugohan58 It's a learning site, not an answer site. Though, we learn and teach through the answering (responding process) :)

@Iventer Do you still have the problem with multiplying the conjugate secx + tanx?

Answer 14

i have heard that, from a friend, thats why i came here to take a look,.......but here i saw an imp member giving all the steps at once....not much thinking will be involved from the asker if all the steps are shown....instead taking one step at a time and and asking what the next step is would help better.....

Answer 15

\[\int\sec\theta\cdot\text d\theta\]\[=\int \sec \theta \times\frac{\sec \theta+\tan \theta}{\sec \theta+\tan \theta}\text d\theta\]\[=\int \frac{\sec^2 \theta+\sec \theta\tan \theta}{\sec \theta+\tan \theta}\text d\theta\]
\[\qquad\qquad\text{let } u={\sec \theta+\tan \theta}\]\[\qquad\qquad \text du=\left(\tan \theta\sec \theta+\sec^2\theta\right)\text d\theta\]\[=\int\frac{\text du}{u}\]\[=\ln u+c\]\[=\ln|\sec \theta+\tan \theta|+c\]

Answer 16

@UnkleRhaukus doing the same thing.....nothing better(except looks)
i am very sorry if i have offended u all
but this is just not what i was told.