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int the third question take tanx common in the numerator, then u get tanx(1+tan^2 x) now u can write it as tanxsec^2 x, nw substitute tanx as t...
in the 4th question, multiply the numerator and denominator by x^2, u get the expression as . (1+a2/x2)/(x2+a4/x2) dx
this is a common form but u might require some manipulations....
in ques 5 put x+1=1/t,u will get through the problem..
i am simplifying the expression
dx= -1/t^2 dt
- dt / t^2(1/t) ( sqrt (1/t^2 - 1/t - 1))
- dt / sqrt(1-t+t^2)
nw it is easy to do...
please rate me if i have helped u in sovilng ur problems
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this expression cannot be integrated...i have tried it several times but unable to do and also i askd ma coaching teachers,they said integrate any expression only if u see it in a book....so can u tell me where is it?
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please someone do this question............................................this question was asked in my coaching examination and this was the way it was given.....
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THE 1st ONE IS DONE BY SUBSTITUTING cot x AS t2, THEN cosec2x dx= -2t dt, sec2x = 1+t4 then the integral becomes [ ][ ] (-2t 2dt)/ 1+t 4 which can be written as [ ] (1-t 2dt)/ 1+t 4 - [ ] (t 2+1dt)/ 1+t 4 i think this becomes easier now.... as tthe above expression is very familiar type of integration.
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two particles 1 & 2 are in shm with same amplitude(A), angular frequency.At time t=0,one is at x=A/2 and the other is at - A/sqrt 2.both are moving in the same direction.when will they collide?
ans: 5T/16 ( T= time period)
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The no. of ways in which three distinct nos in an increasing AP can be selected from the set {1,2,3,.......,24} is.....
answer is :
132
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two players 'a' n 'b' play a series of 2n games.Each game can result in a win or loss for 'a'.The total no. of ways in which 'a' can win the series of game is........
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find the no. of positive integral solutions of
x1x2x3x4x5 = 1050
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hey sorry the figures are placed in the wrong order.
the one that is below is the 1st fig. n the one above is the 2nd fig.
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hey i am assuming that the mass of the sphere is 'm' n that of wedge is 'M' because the image isnt clear.
now the situation b4 the collision is in fig. 1
after collision is in fig. 2
now writing the equations...
(V2/2 + V1 3/2) / Vo/2 = e
and the second eqn is the impulse eqn...
-mVo/2 + F dt = mV2/2
and
Fdt cos 30 = MV1  F dt = 2MV1
please solve the two eqns and reply if the answer u get is correct....
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is it X=A sin ( t +  /6) ?
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moment of intertia of cube
Icm= ma^2/6
about axes perpendicular to every face passin through the centre of mass.......
the axes matter in case of a cuboid because there length,breadth and height are all different....
coz the moi(abt centre of mass) for a cuboid is
Ih = m(w^2 + l^2)/12
Iw = m(h^2 + l^2)/12
Il = m(h^2 + w^2)/12
so as length breadth n height r same so it doesnt matter.....
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hw much time wil a seconds pendulum gain or lose in a day if its lenght is increased by 2%?
answer : a loss of 864 secs
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2 particles execute shm of same amplitude and frequency in parallel lines.they pass one another wen moving in opposite direction each time wen their displacement is half the amplitude.wat is the phase difference between them?
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for the upper block....
limiting condition
25=t+f1 where f1 is the frictional force bw the 2 blocks
for the lower block
limiting condition
t= f1 + f2
f1= *1*10 and f2= * 3 * 10
solving the above eqns..
= 0.5
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since the acceleration of the centre of mass is zero hence its displacement is also zero
thus
mx=m(x-R) where x is the displacement of the tude
we get x=R/2
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pls let me know that if we are to write wbjee2008, then is it necessary to be a resident of west bangal??? i m in delhi n have never lived there so am i eligible???
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PLEASE LET ME KNOW THE WAY OF CONVERTING
ethanol to but1yne...
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THIS IS DONE BY SUBSTITUTING tanx AS t2, THEN sec2x dx=2t dt, sec2x = 1+t4 then the integral becomes [ ] [ ] (2t 2dt)/ 1+t 4 which can be written as [ ] (t 2+1dt)/ 1+t 4 + [ ] (t 2-1dt)/ 1+t 4 i think this becomes easier now.... as tthe above expression is very familiar type of integration. ALL THE BEST
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