Question

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

Answer 1

`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)}`