Given #r=cos theta - sin theta/2-3theta#
Area #A=1/2int_(theta_1)^(theta_2) r^2*d theta#
#A=1/2int_(0)^(pi/2) (cos theta - sin theta/2-3theta)^2*d theta#
#A=1/2int_(0)^(pi/2) (cos^2 theta + sin^2 theta/4+9theta^2-sin theta cos theta+3theta*sin theta-6theta*cos theta)*d theta#
Using Trigonometric identities
#sin^2 A=(1-cos 2A)/2# and #cos^2 A=(1+cos 2A)/2#
Also Integration by parts for
#int theta*sin theta* d theta=-theta*cos theta-int (-cos theta)d theta#
#int theta*sin theta* d theta=-theta*cos theta+sin theta#
Also Integration by parts for
#int theta*cos theta* d theta=theta*sin theta-int (sin theta)d theta#
#int theta*cos theta* d theta=theta*sin theta+cos theta#
#A=1/2 int_(0)^(pi/2) (1/2+1/2cos 2theta+1/8-1/8cos 2theta+9theta^2-sin theta*cos theta+3thetasin theta-6theta*cos theta)*d theta #
#A=1/2 int_(0)^(pi/2)(5/8+3/8cos 2theta+9theta^2-sin theta*cos theta+3thetasin theta-6theta*cos theta)*d theta #
#A=1/2*[(5theta)/8+3/16sin 2theta+3theta^3-1/2*sin^2 theta-3theta*cos theta+3*sin theta-6theta*sin theta-6cos theta]_0^(pi/2)#
#A=1/2[5/8(pi/2)+3/16sin (2(pi/2))+3(pi/2)^2-1/2sin^2 (pi/2)-3*pi/2*cos (pi/2)+3sin(pi/2)-6(pi/2)sin (pi/2)-6cos (pi/2)-(5/8(0)+3/16sin (2(0))+3(0)^2-1/2sin^2 (0)-3*0*cos (0)+3sin(0)-6(0)sin (0)-6cos (0)]#
#A=1/2[5pi/16+0+3pi^3/8-1/2*1^2-0+3(1)-6pi/2-0-(0+0+0-0-0+0-0-6)]#
#A=1/2(5pi/16+3pi^3/8-1/2+3-3pi+6)#
#A=1/2((6pi^3+136-43pi)/32)#
#A=5.8421617492953" "#square units
God bless....I hope the explanation is useful.
