From the reference Arc Length with Polar Coordinates
#L = int_alpha^ beta sqrt(r^2+((dr)/(d theta))^2)d theta#
Given: #r=2sin(theta) -3theta, alpha = (-3pi)/8, and beta=(7pi)/8#
Square the function:
#r^2=4sin^2(theta) -12thetasin(theta)+9theta^2#
Compute the derivative of the function:
#(dr)/(d theta) = 2cos(theta)-3#
Square the derivative:
#((dr)/(d theta))^2 = 4cos^2(theta)-12cos(theta) + 9#
Compute the argument under the radical:
#r^2+((dr)/(d theta))^2 = 4sin^2(theta) -12thetasin(theta)+9theta^2+4cos^2(theta)-12cos(theta) + 9 #
Use the identity #sin^2(theta)+cos^2(theta) = 1#:
#r^2+((dr)/(d theta))^2 = 4(1) -12thetasin(theta)+9theta^2-12cos(theta) + 9 #
Combine like terms:
#r^2+((dr)/(d theta))^2 = -12thetasin(theta)+9theta^2-12cos(theta) + 13#
Substitute into the integral:
#L = int_(-3pi/8)^(7pi/8) sqrt(-12thetasin(theta)+9theta^2-12cos(theta) + 13)d theta#
I used WolframAlpha to evaluate the integral:
#L = int_(-3pi/8)^(7pi/8) sqrt(-12thetasin(theta)+9theta^2-12cos(theta) + 13)d theta ~~ 12.6829#
