What is the area under the polar curve $f (θ) = θ sin (- θ) + 2 cot (\frac{7 θ}{8})$ $f(theta) = thetasin(-theta )+2cot((7theta)/8)$ over $[\frac{π}{4}, \frac{5 π}{6}]$ $[pi/4,(5pi)/6]$ ?

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What is the area under the polar curve $f (θ) = θ sin (- θ) + 2 cot (\frac{7 θ}{8})$ $f(theta) = thetasin(-theta )+2cot((7theta)/8)$ over $[\frac{π}{4}, \frac{5 π}{6}]$ $[pi/4,(5pi)/6]$ ?

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Answer 1 · 2018-03-29T05:20:12

$\int_{\frac{π}{4}}^{\frac{5 π}{6}} f (θ) d θ = \int_{\frac{π}{4}}^{\frac{5 π}{6}} (θ sin (- θ) + 2 cot (\frac{7 θ}{8})) d θ = (\frac{1}{\sqrt{2}} - \frac{1}{2}) - (5 \frac{\sqrt{3}}{12} + \frac{1}{4 \sqrt{2}} π) + \frac{16}{7} \times log ⎧ ⎪ ⎨ ⎪ ⎩ \frac{{sin}^{\frac{6}{5}} (\frac{35 π}{6})}{{sin}^{4} (\frac{7 π}{4})} ⎫ ⎪ ⎬ ⎪ ⎭$ $int_(pi/4)^((5pi)/6)f(theta)d theta=int_(pi/4)^((5pi)/6)(thetasin(-theta)+2cot((7theta)/8))d theta=(1/sqrt2-1/2)-(5sqrt3/12+1/(4sqrt2)pi)+16/7xxlog{sin^(6/5)((35pi)/6)/sin^4((7pi)/4)}$

Explanation:

$f (θ) = θ sin (- θ) + 2 cot (\frac{7 θ}{8})$ $f(theta)=thetasin(-theta)+2cot((7theta)/8)$

$\int_{\frac{π}{4}}^{\frac{5 π}{6}} f (θ) d θ = \int_{\frac{π}{4}}^{\frac{5 π}{6}} (θ sin (- θ) + 2 cot (\frac{7 θ}{8})) d θ$ $int_(pi/4)^((5pi)/6)f(theta)d theta=int_(pi/4)^((5pi)/6)(thetasin(-theta)+2cot((7theta)/8))d theta$

Applying sum rule

$\int_{\frac{π}{4}}^{\frac{5 π}{6}} (θ sin (- θ) + 2 cot (\frac{7 θ}{8})) d θ = \int_{\frac{π}{4}}^{\frac{5 π}{6}} θ sin (- θ) d θ + \int_{\frac{π}{4}}^{\frac{5 π}{6}} 2 cot (\frac{7 θ}{8}) d θ$ $int_(pi/4)^((5pi)/6)(thetasin(-theta)+2cot((7theta)/8))d theta=int_(pi/4)^((5pi)/6)thetasin(-theta)d theta+int_(pi/4)^((5pi)/6)2cot((7theta)/8)d theta$

$\int_{\frac{π}{4}}^{\frac{5 π}{6}} θ sin (- θ) d θ = \int_{\frac{π}{4}}^{\frac{5 π}{6}} θ (- sin θ d θ)$ $int_(pi/4)^((5pi)/6)thetasin(-theta)d theta=int_(pi/4)^((5pi)/6)theta(-sinthetad theta)$
Integrating by parts
$u = θ$ $u=theta$
$\frac{d u}{d} θ = 1$ $(du)/d theta=1$
$d u = d θ$ $du=d theta$

$d v = - sin θ d θ$ $dv=-sinthetad theta$

$\int d v = \int (- sin θ) d θ$ $intdv=int(-sintheta)d theta$
$v = cos θ$ $v=costheta$

$\int u d v = u v - \int v d u$ $intudv=uv-intvdu$

Substituting

$\int θ (- sin θ d θ) = θ cos θ - \int cos θ d θ$ $inttheta(-sinthetad theta)=thetacostheta-intcosthetad theta$

$\int θ (- sin θ d θ) = θ cos θ - sin θ$ $inttheta(-sinthetad theta)=thetacostheta-sintheta$

$\int_{\frac{π}{4}}^{\frac{5 π}{6}} θ sin (- θ) d θ = {θ cos θ - sin θ}_{\frac{π}{4}}^{\frac{5 π}{6}}$ $int_(pi/4)^((5pi)/6)thetasin(-theta)d theta={thetacostheta-sintheta}_(pi/4)^((5pi)/6)$

$= \frac{5 π}{6} cos (\frac{5 π}{6}) - (\frac{π}{4}) cos (\frac{π}{4}) - (sin (\frac{5 π}{6}) - sin (\frac{π}{4}))$ $=(5pi)/6cos((5pi)/6)-(pi/4)cos(pi/4)-(sin((5pi)/6)-sin(pi/4))$

$= \frac{5 π}{6} \times (- \frac{\sqrt{3}}{2}) - \frac{π}{4} \times \frac{1}{\sqrt{2}} - (\frac{1}{2} - \frac{1}{\sqrt{2}})$ $=(5pi)/6xx(-sqrt3/2)-pi/4xx1/sqrt2-(1/2-1/sqrt2)$

Rearranging

$\int_{\frac{π}{4}}^{\frac{5 π}{6}} θ sin (- θ) d θ = (\frac{1}{\sqrt{2}} - \frac{1}{2}) - (5 \frac{\sqrt{3}}{12} + \frac{1}{4 \sqrt{2}} π)$ $int_(pi/4)^((5pi)/6)thetasin(-theta)d theta=(1/sqrt2-1/2)-(5sqrt3/12+1/(4sqrt2)pi)$

$\int_{\frac{π}{4}}^{\frac{5 π}{6}} 2 cot (\frac{7 θ}{8}) d θ = {⎧ ⎪ ⎨ ⎪ ⎩ \frac{2 (log sin (\frac{7 θ}{8}))}{\frac{7 θ}{8}} ⎫ ⎪ ⎬ ⎪ ⎭}_{\frac{π}{4}}^{\frac{5 π}{6}}$ $int_(pi/4)^((5pi)/6)2cot((7theta)/8)d theta={(2(logsin((7theta)/8)))/((7theta)/8)}_(pi/4)^((5pi)/6)$

$= \frac{16}{7} \times {\frac{1}{\frac{5 π}{6}} log sin (7 \times \frac{5 π}{6}) - \frac{1}{\frac{π}{4}} log sin (7 \times (\frac{π}{4})}$ $=16/7xx{1/((5pi)/6)logsin(7xx(5pi)/6)-1/(pi/4)logsin(7xx(pi/4)}$

$= \frac{16}{7 π} \times {\frac{6}{5} \times log sin (\frac{35 π}{6}) - 4 \times log sin (\frac{7 π}{4})}$ $=16/(7pi)xx{6/5xxlogsin((35pi)/6)-4xxlogsin((7pi)/4)}$

$= \frac{16}{7 π} \times {{log sin}^{\frac{6}{5}} (\frac{35 π}{6}) - {log sin}^{4} (\frac{7 π}{4})}$ $=16/(7pi)xx{logsin^(6/5)((35pi)/6)-logsin^4((7pi)/4)}$

$= \frac{16}{7} \times log ⎧ ⎪ ⎨ ⎪ ⎩ \frac{{sin}^{\frac{6}{5}} (\frac{35 π}{6})}{{sin}^{4} (\frac{7 π}{4})} ⎫ ⎪ ⎬ ⎪ ⎭$ $=16/7xxlog{sin^(6/5)((35pi)/6)/sin^4((7pi)/4)}$

$\int_{\frac{π}{4}}^{\frac{5 π}{6}} 2 cot (\frac{7 θ}{8}) d θ = \frac{16}{7} \times log ⎧ ⎪ ⎨ ⎪ ⎩ \frac{{sin}^{\frac{6}{5}} (\frac{35 π}{6})}{{sin}^{4} (\frac{7 π}{4})} ⎫ ⎪ ⎬ ⎪ ⎭$ $int_(pi/4)^((5pi)/6)2cot((7theta)/8)d theta=16/7xxlog{sin^(6/5)((35pi)/6)/sin^4((7pi)/4)}$

$\int_{\frac{π}{4}}^{\frac{5 π}{6}} θ sin (- θ) d θ + \int_{\frac{π}{4}}^{\frac{5 π}{6}} 2 cot (\frac{7 θ}{8}) d θ = (\frac{1}{\sqrt{2}} - \frac{1}{2}) - (5 \frac{\sqrt{3}}{12} + \frac{1}{4 \sqrt{2}} π) + \frac{16}{7} \times log ⎧ ⎪ ⎨ ⎪ ⎩ \frac{{sin}^{\frac{6}{5}} (\frac{35 π}{6})}{{sin}^{4} (\frac{7 π}{4})} ⎫ ⎪ ⎬ ⎪ ⎭$ $int_(pi/4)^((5pi)/6)thetasin(-theta)d theta+int_(pi/4)^((5pi)/6)2cot((7theta)/8)d theta=(1/sqrt2-1/2)-(5sqrt3/12+1/(4sqrt2)pi)+16/7xxlog{sin^(6/5)((35pi)/6)/sin^4((7pi)/4)}$

$\int_{\frac{π}{4}}^{\frac{5 π}{6}} f (θ) d θ = \int_{\frac{π}{4}}^{\frac{5 π}{6}} (θ sin (- θ) + 2 cot (\frac{7 θ}{8})) d θ = (\frac{1}{\sqrt{2}} - \frac{1}{2}) - (5 \frac{\sqrt{3}}{12} + \frac{1}{4 \sqrt{2}} π) + \frac{16}{7} \times log ⎧ ⎪ ⎨ ⎪ ⎩ \frac{{sin}^{\frac{6}{5}} (\frac{35 π}{6})}{{sin}^{4} (\frac{7 π}{4})} ⎫ ⎪ ⎬ ⎪ ⎭$ $int_(pi/4)^((5pi)/6)f(theta)d theta=int_(pi/4)^((5pi)/6)(thetasin(-theta)+2cot((7theta)/8))d theta=(1/sqrt2-1/2)-(5sqrt3/12+1/(4sqrt2)pi)+16/7xxlog{sin^(6/5)((35pi)/6)/sin^4((7pi)/4)}$

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What is the area under the polar curve $f (θ) = θ sin (- θ) + 2 cot (\frac{7 θ}{8})$ $f(theta) = thetasin(-theta )+2cot((7theta)/8)$ over $[\frac{π}{4}, \frac{5 π}{6}]$ $[pi/4,(5pi)/6]$ ?

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