integral sin(x)*ln(sin(x)) dx
using integration by parts with
u=ln(sin(x)) and dv=sin(x) dx then we have
du=cot(x) dx and v=cos(x) thus we have
integral sin(x)*ln(sin(x)) dx=
cos(x)*ln(sin(x))+integral cos(x)*cot(x) dx=
cos(x)*ln(sin(x))+integral cos^2(x)/sin(x) dx
now cos^2(x)=1sin^2(x) so we now have
cos(x)*ln(sin(x))+integral (1sin^2(x))/sin(x) dx=
cos(x)*ln(sin(x))+integral csc(x)sin(x) dx=
cos(x)*ln(sin(x))ln(csc(x)+cot(x))+cos(x)
evaluating this on the interval 0 to 1 we get
ln(2)1 or approximately 0.306853 which agrees with Charlie's numeric solution
Edited on May 19, 2009, 1:06 pm

Posted by Daniel
on 20090519 13:04:15 