Question

guys, let's solve :D

\[\int\limits_{}(\sec^2xdx)/\sqrt{4-tg^2x}\]

Answer 1

solve this urself :P:P
i'm too busy reading jokes and chatting :P:P
do not disturb :P

Answer 2

:/

Answer 3

Me too...

Answer 4

You could use trig identities.

Answer 5

Einstein....wht i wrote earlier means: i don't know how to solve this :'( and i feel so ignorant :(:(
sorry

Answer 6

:( okay :D

Answer 7

i would start by setting tg^2 = k

so you get

sec^x/sqrt(4-kx)

Just looks nicer.

Answer 8

k(x) you mean?
it's a function

Answer 9

Do you remember this identity?:\[\sec^2x=1+\tan^2x\]

Answer 10

yeah I remember

Answer 11

1+tan^2(x) / sqrt(4−tg^2(x))

Answer 12

* dx

Answer 13

Mmm it look even worst.

Answer 14

Yeah,, Can't we make it look like integral of tan^m(x)sec^n(x)dx
it would be easier

Answer 15

Why is tg^2 a function of x?

Answer 16

Lets try u = tan x, so du = sec^2xdx

Answer 17

tan(x), tan^2(x)
tan is a function
so you can't say that tan^2 is a constant like k, like 2, you can't send it out the integral... because it's a fuction, tan(x)

Answer 18

\[\int\limits\frac{du}{\sqrt{4-u^2} }\]

Answer 19

yahh, but we should get rid of this squareroot

Answer 20

I never said tan(x) is a constant

Answer 21

but you tried to do t * x

Answer 22

you took tan as a number, or a variable

Answer 23

Wait. I never took tan as a number. are you saying t=tan?

Answer 24

you wrote kx => k * x ==> tan^2 * x

Answer 25

Where does the tan(x) come from?

Answer 26

tan is a function...

Answer 27

that's why it has an y value for every other x values. beucase it's y = tan(x)

Answer 28

what is g? and what is t?

Answer 29

I was treating them as constants

Answer 30

tg= tan

Answer 31

yh, use substitution method

Answer 32

... are you kidding me? How should we know tg = tan? I thought they were constants

Answer 33

what do you mean?
so what is tg^2?

Answer 34

x.O tg= tan, it sometimes used like that

Answer 35

tg^2(x) == tan^2(x)

Answer 36

ok then :/
starting over...

Answer 37

@slaaiba, Its very common to use tan instead of tg

Answer 38

woah, I've never heard of that notation in my life :/ guess you learn something new everyday

Answer 39

It should be a fairly simple trig substitution then

Answer 40

Yahh, I generally learn, and as No-data said, it's very common to use tg

Answer 41

TANgent line.

Answer 42

right, you just forgot the dx

Answer 43

Can this work:

u = tan x
du = sec^2 x dx

so it's integral of du/sqrt(4-u^2)

Answer 44

\[\int\frac{\sec^2xdx}{\sqrt{4-\tan^2x}}\]

Answer 45

yahh

Answer 46

sure does :)

Answer 47

it's what No-data did slaaibak

Answer 48

Then use the arcsine function

Answer 49

Oh, I didn't see, mybad!

Answer 50

No, it's not that easy slaaibak

Answer 51

trig sub u=2sin(theta), yes?

Answer 52

Isn't \[\int\limits {1\over \sqrt{4-u^2}} du = \arcsin({u \over 2}) + C ?\]

Answer 53

Yes @slaaibak.

Answer 54

yes arcsin(tgx/2) + C , right answer but how? :D

Answer 55

So then arcsin(tanx/2) + C

Answer 56

The integral \(\int \frac{1}{\sqrt{1-u^2}} du=\sin^{-1}u+c.\)

Answer 57

@slaaibak Sorry, you were right

Answer 58

Ohhh, now I got it

Answer 59

If you don't know that formula, and you want to derive it yourself, you can use a trig substitution \(u=2\sin z\), and it will lead you to the same result.

Answer 60

as TurningT just said.

Answer 61

Ok :D

Answer 62

\[\int\frac{du}{\sqrt{4-u^2}}\]\[u=2\sin\theta\to du=2\cos\theta d\theta\]\[\int\frac{\cos\theta}{\sqrt{1-\sin^2\theta}}=\int\sec\theta d\theta=
\ln|\sec \theta+\tan \theta|+C\]then run the whole thing in reverse to get back to x

Answer 63

or wait, I messed up...

Answer 64

should have been 1, not sec(theta)

below the line it would have been cos(theta), not cos^2(theta)

Answer 65

yeah, oops, I forgot to take the square root on the bottom...

Answer 66

just integral d(theta)

Answer 67

\[\int\limits d\theta\]

Answer 68

yeah

Answer 69

Yep!:D
\[=\int d\theta=..\]

Answer 70

\[\int\frac{du}{\sqrt{4-u^2}}\]\[u=2\sin\theta\to du=2\cos\theta d\theta\]\[\int\frac{\cos\theta d\theta}{\sqrt{1-\sin^2\theta}}=\int d\theta =\theta+C\]then run the whole thing in reverse to get back to x

Answer 71

I just like practicing the Latex...

Answer 72

That was typed quite fast

Answer 73

I think he copied pasted somehow :P

Answer 74

I saved it just in case I messed up, and just had to change parts...

Answer 75

I knew it! :P

Answer 76

haha, well played mrmath

Answer 77

lol! I should start doing that. Sometimes, I miss up and have to start all over again!

Answer 78

Yeah, before my pc formatted, I had every question I asked ( as you write an equation when you click on the equation box) in a word folder, but now they ar egone :/