Question

need help on the attachment

Answer 1

Can we possible re write tan^2(x) as someting else, perhaps using an trig identity

Answer 2

I think 1+sec^2=tan^2, but then you will have an integral (sec^(2)x-1)secx

Answer 3

I'm trying it right now, but I'm having a little trouble

Answer 4

also what i the derivative of sec(x)

Answer 5

yeah I tried all day I still don't know how to work it, maybe integration by parts

Answer 6

I'm just failed trying that too, lol!

Answer 7

i don't think this is that easy. you can rewrite in terms of secant alone

Answer 8

either way we need the integral of sec^3x

Answer 9

or maybe splitting the tan^2(x) into separate part like tanx tanx * secx which then maybe integration by parts

Answer 10

I just treid it, you need integral of sec^3x

Answer 11

or tanx

Answer 12

look in the book for those formulas that reduce like the ones for sine and cosine. i will look too

Answer 13

ok

Answer 14

well this is just cheating from the back of the book, but in my copy of stewart it says
\[\int\sec^3(x)=\frac{1}{2}\sec(x)\tan(x)+\frac{1}{2}\ln(\sec(x)+\tan(x)|\]

Answer 15

table of integrals #71

Answer 16

I see how \[\int\limits \sec x=\ln \left| \sec x +\tan x \right|+C\]but, the rest of it...

Answer 17

could you show me how to work the rest because my brain is fried right now

Answer 18

???

Answer 19

here's the integral of sec^3x:
http://www.wolframalpha.com/input/?i=integral+sec%5E3x
I don't think I follow it though...

Answer 20

you see how it comes to integral of (sex^3x+secx)dx right?
well here it is anyway.

Answer 21

http://www.wolframalpha.com/input/?i=integral+tan%5E2x+secx+dx

Answer 22

I can derive integral secxdx if you don;t know it, you want me to show you?

Answer 23

yes

Answer 24

the trick is to multiply top and bottom by secx+tanx:\[\int\limits \sec x dx=\int\limits \sec x{\sec x+\tan x \over \sec x+\tan x}dx\]\[=\int\limits {\sec^2x+\sec x \tan x \over \sec x+\tan x}dx=\ln \left| \sec x+\tan x \right|+C\]

Answer 25

mathframalpha did not show the steps on working out this problem

Answer 26

wolfram you mean?
it did for me... weird
try again maybe?
http://www.wolframalpha.com/input/?i=integral+sec%5E3x
you did click "show steps" right?

Answer 27

There work on the website is kind of confusing to follow because their are using to many different variable. Also could you do the other side too?

Answer 28

you mean \[\int\limits \sec^3x dx\]?

Answer 29

yes

Answer 30

No, I don't know how to do it offhand like the other, but following Wolfram's work it uses a formula called the "reduction formula":\[\int\limits \sec^mx dx={\sin x \sec^{m-1}x \over m-1}+{m-2\over m-1}\int\limits \sec^{m-2}x dx\]In our case m=3.

I have never seen this formula before, by the way.

Answer 31

so what would this look like connecting with the other term of secx

Answer 32

Here's the whole thing on Wolfram:
http://www.wolframalpha.com/input/?i=integral+tan%5E2x+sec+x
note the the log is actually natural log (ln)

Answer 33

I don't get the 5th step answer

Answer 34

still need help

Answer 35

what line? \[\int\limits \sec^3x dx+\ln \left| \sec x+\tan x \right|\]?
or the one after it?

Answer 36

the 5 line on wolframalpha link

Answer 37

does it say whay I posted above? or is it the line after?

Answer 38

what*

Answer 39

yes it what you posted above, and the line after it that gives the solution for it, I just don't get how they came up with the solution for it

Answer 40

The integral of sec^3xdx you mean, because I proved how we got the right part already.
Like i say it uses the reduction formula which I posted above with m=3
I'll walk you through it...

Answer 41

ok, I just don't get how the integral of sec^2xdx could give you what on the bottom,
(1/2)tanxsecx+(1/2)integral sec(x)dx-ln(tan(x)+sec(x))

Answer 42

oops I meant sec^3xdx

Answer 43

\[\int\limits \sec^mxdx={\sin x \sec^{m-1}x \over m-1}+{m-2\over m-1}\int\limits \sec^{m-2}dx\]m=3 gives\[\int\limits \sec^3xdx={\sin x \sec^2x \over2}+{1\over2}\int\limits \sec xdx\]...

Answer 44

since \[\sec x={1\over \cos x}\]\[\sec^2={\sec x \over \cos x}\]so\[{\sin x \sec^2x \over 2}={1\over2}\tan x \sec x\]the integral on the right we already know how to do...

Answer 45

make sense?

Answer 46

yes, but could you also explain why ln(tanx+secx)

Answer 47

is there?

Answer 48

I posted that above! look back up for it.

Answer 49

here it is again:
the trick is to multiply top and bottom by secx+tanx:
\[∫\sec xdx=∫\sec x{\sec x+\tan x \over \sec x+\tan x}dx\]
\[=∫{\sec^2x+\sec x \tan x \over \sec x+\tan x}dx=\ln|\sec x+\tan x|+C\]

Answer 50

why is there a negative on their answer for wolframealpha?

Answer 51

-ln(tanx+secx)

Answer 52

you need to follow the proof more closely, it's all there...
I'll start from the beginning, but try to look for the reason in the meantime to save me the effort please.

Answer 53

ok, sorry

Answer 54

\[\int\limits \tan^2x \sec xdx=\int\limits(\sec^2x-1)\sec xdx=\int\limits \sec^3x-\sec xdx\]\[=\int\limits \sec^3xdx-\int\limits \sec xdx=\int\limits \sec^3xdx-\ln \left| \sec x+\tan x \right|\]I hope you see now. From here we do the integral of sec^3x as shown above.

Answer 55

also on the proof above where did the sinx come from

Answer 56

???

Answer 57

I have no idea, that is part of the "reduction formula" which is used in the proof on Wolfram. As I said I have never seen it before.

Answer 58

ok

Answer 59

I'm sleepy...

Hope I helped a little, but it is a tough integral; even satellite couldn't do it.

G'night!

Answer 60

Thanks!