What is the area under the polar curve f(theta) = thetasin(-theta )+2cot((7theta)/8) over [pi/4,(5pi)/6]?

1 Answer
Mar 29, 2018

int_(pi/4)^((5pi)/6)f(theta)d theta=int_(pi/4)^((5pi)/6)(thetasin(-theta)+2cot((7theta)/8))d theta=(1/sqrt2-1/2)-(5sqrt3/12+1/(4sqrt2)pi)+16/7xxlog{sin^(6/5)((35pi)/6)/sin^4((7pi)/4)}

Explanation:

f(theta)=thetasin(-theta)+2cot((7theta)/8)

int_(pi/4)^((5pi)/6)f(theta)d theta=int_(pi/4)^((5pi)/6)(thetasin(-theta)+2cot((7theta)/8))d theta

Applying sum rule

int_(pi/4)^((5pi)/6)(thetasin(-theta)+2cot((7theta)/8))d theta=int_(pi/4)^((5pi)/6)thetasin(-theta)d theta+int_(pi/4)^((5pi)/6)2cot((7theta)/8)d theta

int_(pi/4)^((5pi)/6)thetasin(-theta)d theta=int_(pi/4)^((5pi)/6)theta(-sinthetad theta)
Integrating by parts
u=theta
(du)/d theta=1
du=d theta

dv=-sinthetad theta

intdv=int(-sintheta)d theta
v=costheta

intudv=uv-intvdu

Substituting

inttheta(-sinthetad theta)=thetacostheta-intcosthetad theta

inttheta(-sinthetad theta)=thetacostheta-sintheta

int_(pi/4)^((5pi)/6)thetasin(-theta)d theta={thetacostheta-sintheta}_(pi/4)^((5pi)/6)

=(5pi)/6cos((5pi)/6)-(pi/4)cos(pi/4)-(sin((5pi)/6)-sin(pi/4))

=(5pi)/6xx(-sqrt3/2)-pi/4xx1/sqrt2-(1/2-1/sqrt2)

Rearranging

int_(pi/4)^((5pi)/6)thetasin(-theta)d theta=(1/sqrt2-1/2)-(5sqrt3/12+1/(4sqrt2)pi)

int_(pi/4)^((5pi)/6)2cot((7theta)/8)d theta={(2(logsin((7theta)/8)))/((7theta)/8)}_(pi/4)^((5pi)/6)

=16/7xx{1/((5pi)/6)logsin(7xx(5pi)/6)-1/(pi/4)logsin(7xx(pi/4)}

=16/(7pi)xx{6/5xxlogsin((35pi)/6)-4xxlogsin((7pi)/4)}

=16/(7pi)xx{logsin^(6/5)((35pi)/6)-logsin^4((7pi)/4)}

=16/7xxlog{sin^(6/5)((35pi)/6)/sin^4((7pi)/4)}

int_(pi/4)^((5pi)/6)2cot((7theta)/8)d theta=16/7xxlog{sin^(6/5)((35pi)/6)/sin^4((7pi)/4)}

int_(pi/4)^((5pi)/6)thetasin(-theta)d theta+int_(pi/4)^((5pi)/6)2cot((7theta)/8)d theta=(1/sqrt2-1/2)-(5sqrt3/12+1/(4sqrt2)pi)+16/7xxlog{sin^(6/5)((35pi)/6)/sin^4((7pi)/4)}

int_(pi/4)^((5pi)/6)f(theta)d theta=int_(pi/4)^((5pi)/6)(thetasin(-theta)+2cot((7theta)/8))d theta=(1/sqrt2-1/2)-(5sqrt3/12+1/(4sqrt2)pi)+16/7xxlog{sin^(6/5)((35pi)/6)/sin^4((7pi)/4)}