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

Mar 29, 2018

${\int}_{\frac{\pi}{4}}^{\frac{5 \pi}{6}} f \left(\theta\right) d \theta = {\int}_{\frac{\pi}{4}}^{\frac{5 \pi}{6}} \left(\theta \sin \left(- \theta\right) + 2 \cot \left(\frac{7 \theta}{8}\right)\right) d \theta = \left(\frac{1}{\sqrt{2}} - \frac{1}{2}\right) - \left(5 \frac{\sqrt{3}}{12} + \frac{1}{4 \sqrt{2}} \pi\right) + \frac{16}{7} \times \log \left\{{\sin}^{\frac{6}{5}} \frac{\frac{35 \pi}{6}}{\sin} ^ 4 \left(\frac{7 \pi}{4}\right)\right\}$

#### Explanation:

$f \left(\theta\right) = \theta \sin \left(- \theta\right) + 2 \cot \left(\frac{7 \theta}{8}\right)$

${\int}_{\frac{\pi}{4}}^{\frac{5 \pi}{6}} f \left(\theta\right) d \theta = {\int}_{\frac{\pi}{4}}^{\frac{5 \pi}{6}} \left(\theta \sin \left(- \theta\right) + 2 \cot \left(\frac{7 \theta}{8}\right)\right) d \theta$

Applying sum rule

${\int}_{\frac{\pi}{4}}^{\frac{5 \pi}{6}} \left(\theta \sin \left(- \theta\right) + 2 \cot \left(\frac{7 \theta}{8}\right)\right) d \theta = {\int}_{\frac{\pi}{4}}^{\frac{5 \pi}{6}} \theta \sin \left(- \theta\right) d \theta + {\int}_{\frac{\pi}{4}}^{\frac{5 \pi}{6}} 2 \cot \left(\frac{7 \theta}{8}\right) d \theta$

${\int}_{\frac{\pi}{4}}^{\frac{5 \pi}{6}} \theta \sin \left(- \theta\right) d \theta = {\int}_{\frac{\pi}{4}}^{\frac{5 \pi}{6}} \theta \left(- \sin \theta d \theta\right)$
Integrating by parts
$u = \theta$
$\frac{\mathrm{du}}{d} \theta = 1$
$\mathrm{du} = d \theta$

$\mathrm{dv} = - \sin \theta d \theta$

$\int \mathrm{dv} = \int \left(- \sin \theta\right) d \theta$
$v = \cos \theta$

$\int u \mathrm{dv} = u v - \int v \mathrm{du}$

Substituting

$\int \theta \left(- \sin \theta d \theta\right) = \theta \cos \theta - \int \cos \theta d \theta$

$\int \theta \left(- \sin \theta d \theta\right) = \theta \cos \theta - \sin \theta$

${\int}_{\frac{\pi}{4}}^{\frac{5 \pi}{6}} \theta \sin \left(- \theta\right) d \theta = {\left\{\theta \cos \theta - \sin \theta\right\}}_{\frac{\pi}{4}}^{\frac{5 \pi}{6}}$

$= \frac{5 \pi}{6} \cos \left(\frac{5 \pi}{6}\right) - \left(\frac{\pi}{4}\right) \cos \left(\frac{\pi}{4}\right) - \left(\sin \left(\frac{5 \pi}{6}\right) - \sin \left(\frac{\pi}{4}\right)\right)$

$= \frac{5 \pi}{6} \times \left(- \frac{\sqrt{3}}{2}\right) - \frac{\pi}{4} \times \frac{1}{\sqrt{2}} - \left(\frac{1}{2} - \frac{1}{\sqrt{2}}\right)$

Rearranging

${\int}_{\frac{\pi}{4}}^{\frac{5 \pi}{6}} \theta \sin \left(- \theta\right) d \theta = \left(\frac{1}{\sqrt{2}} - \frac{1}{2}\right) - \left(5 \frac{\sqrt{3}}{12} + \frac{1}{4 \sqrt{2}} \pi\right)$

${\int}_{\frac{\pi}{4}}^{\frac{5 \pi}{6}} 2 \cot \left(\frac{7 \theta}{8}\right) d \theta = {\left\{\frac{2 \left(\log \sin \left(\frac{7 \theta}{8}\right)\right)}{\frac{7 \theta}{8}}\right\}}_{\frac{\pi}{4}}^{\frac{5 \pi}{6}}$

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

$= \frac{16}{7 \pi} \times \left\{\frac{6}{5} \times \log \sin \left(\frac{35 \pi}{6}\right) - 4 \times \log \sin \left(\frac{7 \pi}{4}\right)\right\}$

$= \frac{16}{7 \pi} \times \left\{\log {\sin}^{\frac{6}{5}} \left(\frac{35 \pi}{6}\right) - \log {\sin}^{4} \left(\frac{7 \pi}{4}\right)\right\}$

$= \frac{16}{7} \times \log \left\{{\sin}^{\frac{6}{5}} \frac{\frac{35 \pi}{6}}{\sin} ^ 4 \left(\frac{7 \pi}{4}\right)\right\}$

${\int}_{\frac{\pi}{4}}^{\frac{5 \pi}{6}} 2 \cot \left(\frac{7 \theta}{8}\right) d \theta = \frac{16}{7} \times \log \left\{{\sin}^{\frac{6}{5}} \frac{\frac{35 \pi}{6}}{\sin} ^ 4 \left(\frac{7 \pi}{4}\right)\right\}$

${\int}_{\frac{\pi}{4}}^{\frac{5 \pi}{6}} \theta \sin \left(- \theta\right) d \theta + {\int}_{\frac{\pi}{4}}^{\frac{5 \pi}{6}} 2 \cot \left(\frac{7 \theta}{8}\right) d \theta = \left(\frac{1}{\sqrt{2}} - \frac{1}{2}\right) - \left(5 \frac{\sqrt{3}}{12} + \frac{1}{4 \sqrt{2}} \pi\right) + \frac{16}{7} \times \log \left\{{\sin}^{\frac{6}{5}} \frac{\frac{35 \pi}{6}}{\sin} ^ 4 \left(\frac{7 \pi}{4}\right)\right\}$

${\int}_{\frac{\pi}{4}}^{\frac{5 \pi}{6}} f \left(\theta\right) d \theta = {\int}_{\frac{\pi}{4}}^{\frac{5 \pi}{6}} \left(\theta \sin \left(- \theta\right) + 2 \cot \left(\frac{7 \theta}{8}\right)\right) d \theta = \left(\frac{1}{\sqrt{2}} - \frac{1}{2}\right) - \left(5 \frac{\sqrt{3}}{12} + \frac{1}{4 \sqrt{2}} \pi\right) + \frac{16}{7} \times \log \left\{{\sin}^{\frac{6}{5}} \frac{\frac{35 \pi}{6}}{\sin} ^ 4 \left(\frac{7 \pi}{4}\right)\right\}$