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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.

5 years ago


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


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

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

          I/pi   =-pilog2





5 years ago

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


replace x by (0+∏)-x

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


I=∏logsinx -I



then using another property of definte integration


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)



final answer

5 years ago

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