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What is the integration of {xlog(sin(x)} in lower limit is o and upper limit is pie


tell me it with proper steps.

6 years ago

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Answers : (2)

										

this q integration is too lenthy , here is the complete solution ...


I = xlog(sinx)dx     .........1                   lim 0 t0 pi


I =(pi-x)logsin(pi-x)dx                    lim 0 to pi                      (using property of integration)


I = -xlog(sinx)dx  + pilog(sinx)dx ................2              lim 0 to pi


adding 1 & 2


2I = pilogsinxdx or


2I/pi = logsinxdx  .....................3           lim 0 to pi    


 


 2Ipi = logsinxdx + logsinpi-x  dx      lim 0 to pi/2                (by property)


 2I/PI = 2logsinx dx ................4      lim 0 to pi/2


 2I/pi = 2logcosx dx ................5   lim 0 to pi/2           (by property)


 adding 4 & 5


 4I/pi =2log(sin2x)/2 dx       lim 0 to pi/2


  2I/pi = logsin2xdx - log2 dx        from 0 to pi/2    


  2I/pi =   I1 - log2dx                      lim 0 to pi


from this eq 6 integrating logsin2x seperately


 I1 = logsin2xdx             lim 0 to pi/2


put 2x =t


 I1 = (logsintdt)/2  lim 0 to pi


now interchanging variable t with x


 I1 = logsinxdx/2            lim 0 to pi ........................7


from eq 3 & 7


I1 = I/pi .........8


putting eq 8 in 6


then


        2I/pi = I/Pi  - log2dx    lim 0 to pi


       I/pi = -(log2)x      lim 0 to pi


          I/pi   =-pilog2


 


 


 


 

6 years ago
										

hi. i'm not writting the integral sign and the limits manage with this




 xlog(sinx)=I


replace x by (0+∏)-x


I=[(0+∏)-x]log sin(∏-x)=(∏-x)log(sinx)


I=∏log(sinx)-xlog(sinx)


I=∏logsinx -I


2I=∏logsinx 


(2/∏)I=∫logsinx=F


then using another property of definte integration


i.e


o  f(x)=o(∏/2) f(x) +o(∏/2) f(2a-x)


F= 0∫?(∏/2)logsinx+ 0(∏/2)logsin(∏-x)=2logsinx


F/2=-(∏/2)log2     (the above one is a standard integral)


F=-∏log2=(2/∏)I


I=-(∏2/2)log2


final answer





6 years ago

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