# What is the net area between f(x) = cotxcscx -sinx and the x-axis over x in [pi/6, (5pi)/8 ]?

Mar 26, 2016

$A = {A}_{1} + {A}_{2} = 2 - \frac{1}{\sin} \left(\frac{5 \pi}{8}\right) + \cos \left(\frac{5 \pi}{8}\right) - \frac{\sqrt{3}}{2} \approx - .331$

#### Explanation:

$A = {\int}_{\frac{\pi}{6}}^{\frac{5 \pi}{8}} \left(\cot x \csc x - \sin x\right) \mathrm{dx}$
${A}_{1} + {A}_{2} = {\int}_{\frac{\pi}{6}}^{\frac{5 \pi}{8}} \cot x \csc x \mathrm{dx} - {\int}_{\frac{\pi}{6}}^{\frac{5 \pi}{8}} \sin x \mathrm{dx}$
${A}_{1} = {\int}_{\frac{\pi}{6}}^{\frac{5 \pi}{8}} \cot x \csc x \mathrm{dx}$
${\int}_{\frac{\pi}{6}}^{\frac{5 \pi}{8}} \cos \frac{x}{\sin} x \cdot \frac{1}{\sin} x \mathrm{dx} = {\int}_{\frac{\pi}{6}}^{\frac{5 \pi}{8}} \cos \frac{x}{\sin} ^ 2 x \mathrm{dx}$

Let $u = \sin x \implies \frac{\mathrm{du}}{\mathrm{dx}} = \cos x$ then rewrite the integral as:
${A}_{1} = \int \frac{\mathrm{du}}{u} ^ 2 = - \frac{1}{u}$ substitute u=sinx
${A}_{1} = - \frac{1}{\sin} x = - {\left[\csc x\right]}_{\frac{\pi}{6}}^{\frac{5 \pi}{8}} = 2 - \frac{1}{\sin} \left(\frac{5 \pi}{8}\right)$

${A}_{2} = {\int}_{\frac{\pi}{6}}^{\frac{5 \pi}{8}} \sin x \mathrm{dx} = {\left[- \cos x\right]}_{\frac{\pi}{6}}^{\frac{5 \pi}{8}}$
${A}_{2} = \cos \left(\frac{5 \pi}{8}\right) - \frac{\sqrt{3}}{2}$

$A = {A}_{1} + {A}_{2} = 2 - \frac{1}{\sin} \left(\frac{5 \pi}{8}\right) + \cos \left(\frac{5 \pi}{8}\right) - \frac{\sqrt{3}}{2} \approx - .331$