TargetIIT

[ss]o[/ss]∫[p]pi/2[/p] dx/ [sqrt]1-ksin[p]2[/p]x[/sqrt]


yeh pleeeeeeeeeez koi banao ....

[hide]kal poori saat nahi so payee theek se.. iske chalte...
have bcome dumb in integration i think :( :([/hide]

[7]

did u notice k cant b more than 1

given integration can b written as......1/[sqrt]m-sin[p]2[/p][/sqrt]x

yes i did noticed that ....

yup ...
tried in that way too...

but then ??

how r u goingto proceed afetr that ?

in dAT way denominator is coming to be root under two quadratic......

arey plz do naa....

ok tell me one thing...

how to integrate this one:

∫dt/ [sqrt]t[p]4[/p]+t[p]2[/p]+1[/sqrt]  ??

fro that traditional method t-1/t =sumthing

we need tdt at the num... or something else...

par yeh nahi banra us technique se [2]

ya got it......


forget above blah blah....


put x=0+pi/2-x;

deno. becomes  [sqrt]1=-mcos[p]2[/p]x[/sqrt];


add th e two  this will give 2I=1/[sqrt]K[/sqrt]SINX.COSX

NO NO AGAIN WRONG........NUMERATOR TOH BHUL HI GAYE...

pehle ques ke bare mein soch raha hoon
for the 2nd one make the whole sq in the demominator
putt[p]2[/p]+1/2=k so
t=[sqrt]k+1/2[/sqrt] so find dt and hence ans

[b][i][u]and wait for some time
and let us think on it[/u][/i][/b]

yeh kaise a gaya ???


mereko toh ni aara !! [12]

hmmmmmmmmmmmm!!

sorry mkagenius k could be any integer

@mani, koi nahi tha ....

NO! k cant be any integer ...

and sky check the ques again kahin k[p]2[/p]to nahin hai

WHY??????????????

sinx bade badoan ki band baja deta hai

@mathie, how can u leave anything ?

leave it... n start trying :P

pehle yeh batao ki 2nd ques samajh aa gaya ??????????

2nd ??

ek hi toh question hai !

ia m saying about the 10nd post not the 1st one

see, [sqrt]1-ksin[p]2[/p]x[/sqrt] = [sqrt]k[/sqrt] [sqrt]1/k - sin[p]2[/p]x[/sqrt]

now in [sqrt]1/k - sin[p]2[/p]x [/sqrt]

1/k - sin[p]2[/p]x >0

if K >1 => 1/k <1

sin[p]2[/p]x ε [0,1]

=> 1/k - sin[p]2[/p]x <0 for some x ...

this is not possible...

so 0<k<1 always..

omg! powercut!! in this crucial time!

plz plz plz solve this ...

hats off to the one who will do it [in advance]

sky ji sinx 0 ho sakta hai
so cant divide it

MY ANSWER WOULD BE  Π/2 SE THODA JYAADA AND Π/2[sqrt]1-K[/sqrt]  SE THODA KAM...

HURRAY
ho gaya
put sinx = t
so sin[p]-1[/p]t=x
hence dx=dt/[sqrt]1-t[p]2[/p][/sqrt]
so the ques becomes
dt/( [sqrt]1-t[p]2[/p][/sqrt]X [sqrt]1-kt[p]2[/p][/sqrt])
i hope now u will be able to solve

WELL DATS WHAT I SAID IN POST NO 8.......


BUT WE R TRYING TO SOLVE IN A SHORTER WAY....


BTW GOOD JOB....

CAN U PROCEED AFTER DAT AND POST THE FULL SOLUTION.....

i didnt saw ur post
but i think this is the only way to proceed

WHILE PROCEEDING DONT FORGET K<1

@SKY EDITING REQUIRED IN POST NO......29.......K CAN BE NEGATIVE......

0 se pi/2 ke beech main sinx -ve??????????????

koi pura solve kar do mere bhai!!!!1

well manipal... uske agey banao na bhai !

yeh toh mujhe question milne ke 3 min baad hi mil gaya tha ....

my doubt [b]is iske agey .... [/b]

hey sky this seems to be legendre's integral or something like dat....its not in the syllabus ....

arey ... may be i also think its not in syll ...

[hide]woh ... ek friend ne banane ko kaha tha...

she is in iit,kgp...they r having mid-sem....so asked to try...

mereko dekhke lega asaan hai... par :'( :'([/hide]

MAKE IT OF FORM∫dx/1-(x) =(sin)-1 (x) so answer is (sin)-1 √k

@skygirl: are you sure the qn needs you to integrate the function. Or are they aksing something else (like its increasing or decreasing??)

no bhaiya... she wanted it to be integrated only....

well take the 1 as cos[p]2[p][/p][/p]x + sin[p]2[/p]x  

then it reduces to

[ss]0[/ss]∫[p]π/2[/p] 1 / √[o]cos[p]2[/p]x + (k-1)sin[p]2[/p]x[/o]

then by properties .
we get

2I  =  [ss]0[/ss]∫∫[p]π/2[/p]  1 / √[o][o](sin[p]2[/p]x + (k-1)cos[p]2[/p]x ) ( cos[p]2[/p]x + (k-1) sin[p]2[/p]x )[/o][/o]

then well divide Nr and Dr by cos[p]2[/p]x then this should be over hopefully. is this ok ?

no :(

i did this too :(

leave it yaar.... i mean if u like doing... then no prbs :)

cant we do

   by

[ss]0[/ss]∫[p]pi/2[/p]dx/[sqrt]1-ksin[p]2[/p]x[/sqrt]=[ss]0[/ss]∫[p]pi/2[/p]dx[sqrt]1-kcos[p]2[/p]x[/sqrt]


take I=[ss]0[/ss]∫[p]pi/2[/p]dx/[sqrt]1-ksin[p]2[/p]x[/sqrt]

      2I= [ss]0[/ss]∫[p]pi/2[/p]dx/[sqrt]1-ksin[p]2[/p]x[/sqrt]+[ss]0[/ss]∫[p]pi/2[/p]dx[sqrt]1-kcos[p]2[/p]x[/sqrt]
Try this manner tell me if ur getting........

can we do in this method sky.....plzz reply.....