This has a very long solution but, I will only include the summary
The formula Area #=1/2int_alpha^beta r^2 d theta#
Let #r=theta^2 - theta* sin(6 theta-pi/12)+cos(12 theta-(5 pi)/3)#
Area #=1/2int_alpha^beta r^2 d theta#
Area #=1/2int_alpha^beta (theta^2 - theta* sin(6 theta-pi/12)+cos(12 theta-(5 pi)/3))^2 d theta#
Expanding the integrand results to
Area #=1/2int_alpha^beta [theta^4 - theta^2* sin^2 (6 theta-pi/12)+cos^2 (12 theta-(5 pi)/3)-2*theta^3*sin(6 theta-pi/12)-2*theta*sin(6 theta-pi/12)*cos(12 theta-(5 pi)/3)+2*theta^2*cos(12 theta-(5 pi)/3)]d theta#
After trigonometric transformation, it follows
let #beta=pi/6# and #alpha=pi/8#
Area #=1/2int_(pi/8)^(pi/6)[theta^4+1/2*theta^2-theta^2/2*cos(12 theta-pi/6)+1/2#
#+1/2*cos(24 theta-(10pi)/3)-2*theta^3*sin(6theta-pi/12)#
#-theta*sin(18 theta-(7pi)/4)+theta*sin(6theta-(19pi)/12)#
#+2*theta^2*cos(12theta-(5pi)/3)] d theta#
After using Integration by Parts to integrate, it follows
Area#=1/2[theta^5/5+theta^3/6-1/24theta^2sin(12theta-pi/6)-theta/144cos(12theta-pi/6)#
#1/1728sin(12theta-pi/6)+1/2theta+1/48sin(24theta-(10pi)/3)#
#+theta^3/3cos(6theta-pi/12)-theta^2/6sin(6theta-pi/12)-theta/18cos(6theta-pi/12)#
#+1/108sin(6theta-pi/12)+theta/18cos(18theta-(7pi)/4)-1/324sin(18theta-(7pi)/4)-theta/6cos(6theta-(19pi)/12)+1/36sin(6theta-(19pi)/12)+theta^2/6sin(12theta-(5pi)/3)+1/36thetacos(12theta-(5pi)/3)-1/432sin(12theta-(5pi)/3)]_(pi/8)^(pi/6)#
Area #=1/2*[0.3085665067-0.2207297217]#
#color(red)("Area") ##color(red)(=0.04391839249" ")#square unit
God bless....I hope the explanation is useful.
