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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. What is the integration of {xlog(sin(x)} in lower limit is o and upper limit is pie tell me it with proper steps.
What is the integration of {xlog(sin(x)} in lower limit is o and upper limit is pie
tell me it with proper steps.
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
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
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
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
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