ppl......i was jus. trying these probs today.....mere palle nahi pad rahe the in many of these....jus try it out....

1) if A and B are the roots of the eqn. tanx = 2x.........find the value of _-1S⁺¹sinAx . sinBx dx  (A,B>0)

2) find _-3S⁺³x⁸{x¹¹} dx (my formula edittor isnt working)

3)the pts represented by vectors a,b,c and d are coplanar and (sinA)a + (2sin2B)b + (3sin3C) -4d =0

then, find the least value of sin²A + sin²2B + sin²3C

4) find the domain of f(x) =log_e(3(pi)x - x² - 2(pi)² - sinx )

try it out and nudge if u get it......

ok pramod6990 this is a very, very,very rough method so please scrutinize my soln properly

the thing in the brackets of the log should be greater than 0 so i put

3piex-x^2-2pie^2-sinx>0so i get x^2-3piex+2pie^2<-sinx so

(x-pie)(x-2pie)<-sinx so (pie-x)(x-2pie)>sinx @

what i told you earlier(-pie,2pie) was wrong i had made a mistake but now if you analize equation@ i get my domain as (pie,2pie)

please tell me if you wan't me to arrive at the domain after analyzing as it is difficult to explain it out here

1)_-1S⁺¹sinAxsinBx dx=₀ s¹2sinAx sinBx dx=₀ s¹cos(A-B)x-cos(A+B)x dx=sin(A-B)x-sin(A+B)x=sin(tanA-tanB/2)-sin(tanA+tanB/2)=2cos(tanA)cos(tanB)=2cos2Acos2B ANS

3)use this result



(Aa + Bb +Cc)² ( A² + B² + C² ) (a² + b² + c² )

sin²A + sin²2B + sin²3C>=((sinA)a + (2sin2B)b + (3sin3C))²/(a²+4b²+9c²)

=> least value =16d²/(a²+4b²+9c²)  ----------

good mtd mudit...although a rough procedure...a nice one......

and akku....superb soln. to that integration one....and about that vectors one.....ur soln is 16d²/(a²+4b²+9c²)......the answer is 8/7{options are 8/7 7/8 1 4}.....(the fact that these are unit vectors is perhaps missed out).....but i am jus. thinking...using this identity(cauchy-schwartz if i amnt mistaken) how do we make use of the fact that these vectors are coplanar.....neways i feel like kicking my self for missing out on that(identity)....neways rated u ppl....