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atleast someone try it!!!!!!!
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A is 1,2-dibromo cyclohexane
B is cyclohexene
C is cyclohexanone
D is cyclohexanol
E is cyclohexene
is my answer correct>>>>>>>>
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put log x=t
dx/x = dt
I =  t dt = t2/2 + c
so log x dx (log x)2
----------- = -------------- + c
x 2
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go to this link ......... i have already solved the problem>>>>>>>
http://www.goiit.com/posts/list/integration-dx-sin-x-5-cos-x-5-integrate-this-25006.htm
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solution for the second problem.............
let I = (secx)^2 dx from 0 to pi/2
---------------------------------
(secx + tanx)^n
put secx + tanx = t.............(i)
dt = (secxtanx + (tanx)^2)dx
dt = secx(secx+tanx)dx
dt/t = secxdx.............(ii)
(secx)^2 - (tanx)^2 = 1
[secx+tanx][secx-tanx]=1
[secx-tanx]=1/t...............(iii)
adding (i) and (iii) . we get
2secx = t+ 1/t
=> secx = (t2 + 1)/2t........(iv)
therefore the integral becomes
I = (t2 + 1)dt from 1 to infinity
------------------
tn * 2t * t
I =1/2[ t-n dt + t-n-2dt ] from 1 to infinity
I = 1/2[ [t1-n / (1-n)] + [ t-n-1 /(-1-n)] ] from 1 to infinity
applying the limits we get........
I = 1/2[ -1/(1-n) + 1/(n+1) ]
I= 1/2 [ 1/(n+1) - 1/ (1-n) ]
taking LCM we get
I = 1* [ 1-n -n -1 ]
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2 * (n+1)*(1-n)
I = 2*n
------------------------
2 * (n+1) * (n-1)
thus we get........
I = n/(n2 - 1)
i hope you are clear with my proof>>>>>>>>>>
please do rate me .......................
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put x+a=t
dx=dt
x-a = t-2a
substituting you will get sin(t-2a)/sint
now expand the numerator sin(t-2a) ........
you will get the answer...........
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if
X = 1 + 1/ 2 + 1/ 3+ 1/ 4+ 1/ 5+ ..................... + 1/ 106 .
find [X] where [ ] : denotes greatest integer function
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visit www.integrals.wolfram.com and type your question....
you will get the answer.......
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the answer is x18=y21=z28 . the proof is .........
let logyx=a, logzy=b, logxz=c .......
given 3,3a,3b,7c are in AP..
6a=3+3b => 2a=b+1..........(i)
6b=3a+7c................(ii)
abc=1.........(iii)
multiply (ii) by ab we get
6ab2=3a2b+7abc
6a(2a-1)2=3a2(2a-1)+7
solving this equation we get a=7/6
substitute a=7/6 in (i) we get b=4/3
substitute values of a and b in (iii) we get c=9/14
therefore
logyx=7/6 => x=y7/6 => x6=y7 => x18=y21
logzy=4/3 => y=z4/3 =>y3=z4 => y21=z28
logxz=9/14 => z=x9/14 => z14=x9 => z28=x18
thus we get x18=y21=z28 ......
answer 2) is correct
please do rate me.......
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i think the product is 3-bromo cyclopent-1-ene
because we get only allylic aubstitution..........
allylic free radical is stabilised by resonance.......
what is the right answer??????
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how will you distinguish between
C6H5-NH-CH3 and CH3-NH-CH3 ??????
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the solution is x=2 ..........
i think no other value will be the solution..........
if you substitute x=2 ..... you will get
(5+2 6)4-3 + (5-2 6)4-3 = 5 + 2 6 + 5 - 2 6 = 10
please rate me if i am correct............
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experts please solve it>>>>>>>>>>>
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let I = dx
----------------------------
(sinx)^5 + (cosx)^5
I = dx
------------------------------------------------------------------------------
(sinx +cosx)(1 - sinxcosx - (sinxcosx)^2)
I = (sinx + cosx)dx
-------------------------------------------------------------------------
( 1 - 2sinxcosx) ( 1 - sinxcosx - (sinxcosx)^2)
now proceed by substituting sinx - cosx = t
squaring you will get
1 - 2sinxcosx=t^2
sinxcosx= 1/2*(1 - t^2)
substitute in the integral and proceed via the method of partial fractions ..............................
please do rate me if you are satisfied with my reply>>>>>>>>>
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i have written the important equations of state which gives the relation between pressure temperature and volume........
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