1[12]
chhichhi bad mistake [2]
but then how to discuss monotonocity ??
o∫pi/2 dx/ √1-ksin2x
yeh pleeeeeeeeeez koi banao ....
kal poori saat nahi so payee theek se.. iske chalte...
have bcome dumb in integration i think :( :(
-
UP 0 DOWN 0 0 69
69 Answers
1yup thts true...................def integral is a constant (if k here is a constant) so its differentiation is 0.
1definite integral ka monotonic behaviour syllabus be kab se aa gaya....
......it willl beeee zerooooooooooo.....i.e f dash k
341What's in the syllabus is this:
Consider f(k) = 0∫π/2 1/√1-k sin2x dx 0<k<1
Discuss monotonic behaviour of f(k) in (0,1)
11pehle ques ke bare mein soch raha hoon
for the 2nd one make the whole sq in the demominator
putt2+1/2=k so
t=√k+1/2 so find dt and hence ans
1ya got it......
forget above blah blah....
put x=0+pi/2-x;
deno. becomes √1=-mcos2x;
add th e two this will give 2I=1/√KSINX.COSX
1fro that traditional method t-1/t =sumthing
we need tdt at the num... or something else...
par yeh nahi banra us technique se [2]
1in dAT way denominator is coming to be root under two quadratic......
11i didnt saw ur post
but i think this is the only way to proceed
1WELL DATS WHAT I SAID IN POST NO 8.......
BUT WE R TRYING TO SOLVE IN A SHORTER WAY....
BTW GOOD JOB....
11HURRAY
ho gaya
put sinx = t
so sin-1t=x
hence dx=dt/√1-t2
so the ques becomes
dt/( √1-t2X √1-kt2)
i hope now u will be able to solve
1MY ANSWER WOULD BE Î /2 SE THODA JYAADA AND Î /2√1-K SE THODA KAM...
1omg! powercut!! in this crucial time!
plz plz plz solve this ...
hats off to the one who will do it [in advance]