Question

answer the question that ill post to down

Answer 1

\[\int\limits_{?}^{?}(\sqrt{tanx})*dx\]

Answer 2

\[\int\limits_{?}^{?}(\sqrt{tanx})*dx\]

Answer 3

\[look \ at \ this\ ]

Answer 4

there is a mistake in u getting the relation between dx and dw....
wen tanx=w^2
tan^2x=w^4......check and re-substitute :)

Answer 5

what about his= if we take tanx= u then u=sinx/cosx then du=cos2x-sin2x*dx ya?

then it will be like that: \[ \int\limits_{}\sqrt{sinx/tanx}*(\cos2x-\sin2x*dx)\]

Answer 6

i thikn this is a betta solition way

Answer 7

O.o how did ya get all thaaaaaaaaaaaat?> O.o

Answer 8

^^

Answer 9

that isn't an answer is that ? :P

Answer 10

no its just the way i dunno if its true or not :P

Answer 11

tanx = t^2
2tdt = sec^2 x dx
dx = 2tdt/(1+t^2)
so u get 2t^2 dt/ (1+t^2)

Answer 12

what is sec Him?

Answer 13

whats t in there??

Answer 14

2t - 2atant
2root(tanx) -2atan(root(tanx))

Answer 15

t is another variable..what you guys call the variable "u"

Answer 16

ahh k ^^

Answer 17

is it fine?

Answer 18

but 2a??

Answer 19

I still didn't get what sec is...:(:(

Answer 20

thats 2arctant

Answer 21

sec means secant (x)

Answer 22

ok i have to go for dinner... :(((( brb
AHH OKAY NOW GOT IT

Answer 23

yay....medal

Answer 24

i think i got it..thnx :):)

Answer 25

thanks much

Answer 26

him! u also made the same mistake!

Answer 27

dx=2tdt/(1+t^4)

Answer 28

uh oh

Answer 29

2t^2/(1+t^4)

Answer 30

wow...i was out for walk and so much of convos taken place... nice :)

Answer 31

so how do u integrate this?

Answer 32

2t^2/(1+t^4)

Answer 33

u mean how i intagrate this?

Answer 34

multiply and divide by t^2

Answer 35

leave the 2 out for a second

Answer 36

now uve got t^4dt/(1+t^4)t^2

Answer 37

now add and subtract one from the denominator

Answer 38

dt/t^2 - dt/(1+t^4)t^2

Answer 39

for the second term there u need to do a partial fraction decompose

Answer 40

so ull get (this is for the second term)

dt/t^2 - t^2 dt/ (1+t^4)

Answer 41

now the second term of this is equal to ur original integral

Answer 42

bt then it doesnt add up....pellet..pellet

Answer 43

any ideas praveen

Answer 44

aj main is bhen**** ko kare bina so oonga nahi

Answer 45

i doubt if u can proceed dat way...
i remember doing these in 11th grade...lost all touch.
write 2t^2=t^2+1 +t^2-1, and then split the integral...
that is::((1+t^2)/1+t^4)+((t^2-1)/1+t^4)
so anyideas from here?

Answer 46

U guys are still working in this 1? WOW!!! :P

Answer 47

this is where m stuck 2

Answer 48

hey i got it
aftr that step i told
divide numerator and denominator by t^2
so denom ull get as t²+1/t²
which is
(t+1/t)² -2 or (t-1/t)²+2
take t+1/t as som x and t-1/t as y
and den simple integ

Answer 49

\[\int_{}\sqrt{tan(x)}\ dx\]
i wonder....
\[z^{2} = tan(x)\ \ \ \ \ x=tan^{-1}(z^2)\]
\[2z = sec(tan^{-1}(z^2))dx\]
\[\frac{2z}{sec(tan^{-1}(z^2))}=dx\]
\[\int_{}\frac{2z^3}{sec(tan^{-1}(z^2))}\ dz\]
\[\int_{}{2z^3cos(tan^{-1}(z^2))}\ dz\]
not much better is it ;)

Answer 50

or is it?.... hmmmm

Answer 51

bt then wht of the numerator man?

Answer 52

lol...korcan , i've given the easiest solution i guess... its pretty simple, follow the steps i told, post if ur facing problems :)

Answer 53

....cept for the \(2z^3\) needs to be cahnged to \(2z^2\) perhaps

Answer 54

done

Answer 55

main duniy ka sabse bada chutiya hoon na

Answer 56

numerator will be 1+1/t^2 and 1-1/t^2 , where there is 1+1/t^2 , substitue t-1/t=x, and where there is 1-1/t^2 substitute t+1/t=y.
about the denominator , its t^2+1/t^2 which can be factorised as (t+1/t)^2 or (t-1/t)^2

Answer 57

yes yes got it

Answer 58

IM ASHAMED OF MYSELF

Answer 59

dude...i guess these guys don't know hindi... :D lol chill saale :)

Answer 60

m still elder lol:)

Answer 61

oh! i don't medals! damn! :D

Answer 62

korcan!angela! come on! spring some booster by giving medals! i've been working on this for 15 mins!

Answer 63

Hahaha...Ok....but....I still don't get it ......I'll ask my teacher to explain that to me :)

Answer 64

im still pretty ashamed of myself...