Area #= 1/2 int r^2 d theta#,
with the given r and #theta# from #pi/8# to #pi/4#
#=1/2 int sin^2(5theta-13/12pi) d theta#, with the limits
#=1/4int (1+cos(10theta-13/6pi) d theta#, with the limits
#1/4[theta+1/10sin(10theta-13/6pi)]#, between #theta = pi/8 and pi/4#
#=1/4(pi/4-pi/8)+1/10(sin(pi/2-pi/6)-sin(-11/12pi))#
#=1/4(pi/8-1/10(sin(pi/3)+sin(pi/12)))#
#=1/4(pi/8-1/10(sqrt3/2+sqrt((1-sqrt3/2)/2)))#.
Note: There are some nuances. In strict mathematical parlance,
#r = sqrt(x^2+y^2)>=0.# Here, the curve is a 5-petal rose, with one
petal created in one period #P =2/5pi#.
Yet, here, r < 0 for part of the given interval. You can see that
#r(pi/8)=-sin(pi/12)<0#. This means, the area comprises areas from
#r_- and r_ +# petals. There are altogether 5 petals from r >=0 and
5 more, for conventionally added r <=0 petals.
Interested readers can ponder over this aspect of the problem.
