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

Mar 8, 2016

$F \left(x\right) = {F}_{1} \left(x\right) - {F}_{2} = {\left[\ln \tan \left(\frac{x}{2}\right) - \sin x + x \cos x\right]}_{\frac{\pi}{6}}^{\frac{5 \pi}{8}}$
$F \left(x\right) \approx .09143$

#### Explanation:

The area under $f \left(x\right) = \csc x - x \sin x$ and the $x$ axis over the interval $\left[\frac{\pi}{6} , \frac{5 \pi}{8}\right]$ is obtained by compute the integral of $f \left(x\right)$,
$F \left(x\right) = {\int}_{\frac{\pi}{6}}^{\frac{5 \pi}{8}} f \left(x\right) \mathrm{dx} = {\int}_{\frac{\pi}{6}}^{\frac{5 \pi}{8}} \left(\csc x - x \sin x\right) \mathrm{dx}$
Apply the Sum Rule:
$F \left(x\right) = {\int}_{\frac{\pi}{6}}^{\frac{5 \pi}{8}} \left(\csc x\right) \mathrm{dx} - {\int}_{\frac{\pi}{6}}^{\frac{5 \pi}{8}} \left(x \sin x\right) \mathrm{dx}$
Let
${F}_{1} \left(x\right) = {\int}_{\frac{\pi}{6}}^{\frac{5 \pi}{8}} \left(\csc x\right) \mathrm{dx}$ ============> (1)
${F}_{2} = {\int}_{\frac{\pi}{6}}^{\frac{5 \pi}{8}} \left(x \sin x\right) \mathrm{dx}$ =============> (2)

Let's integrate $\textcolor{red}{\left(1\right)}$ applying integral substitution:
Let u= tan(x/2); then $\csc \left(x\right) = \setminus \frac{1 + {u}^{2}}{2 u}$
and $\mathrm{dx} = \setminus \frac{2}{1 + {u}^{2}} \mathrm{du}$
${I}_{1} \left(u\right) = \setminus \int \setminus \frac{1 + {u}^{2}}{2 u} \setminus \frac{2}{1 + {u}^{2}} \mathrm{du} = \setminus \int \setminus \frac{1}{u} \mathrm{du}$
I_1(u)= |ln(u)|; substituting $u = \tan \left(\frac{x}{2}\right)$
${F}_{1} = \left[| \ln \left(\tan \left(\frac{x}{2}\right)\right) |\right]$

Now let integrate $\textcolor{b l u e}{\left(2\right)}$ applying integration by parts. Recall from the product rule of derivative of 2 function f and g
$\left(f \cdot g\right) ' = f ' g + g ' f$ integrate both side and you end with
$\int \left(f \cdot g\right) ' \mathrm{dx} = \int \left(f ' g + g ' f\right) \mathrm{dx}$
$\left(f \cdot g\right) = \int \left(f ' g + g ' f\right) \mathrm{dx}$ we can rewrite this as:
$\textcolor{p u r p \le}{\int \left(g ' f\right) \mathrm{dx} = \left(f \cdot g\right) - \int \left(f ' g\right) \mathrm{dx}}$ ======>(3)
From what we derived in (3) and letting:
u = x; (du)/dx = u' = 1
v' = sinx; v = int sinx dx = -cosx
$\textcolor{p u r p \le}{\int u v ' \mathrm{dx} = u v - \int u ' v \mathrm{dx}}$
$\int x \sin x \mathrm{dx} = x \left(- \cos x\right) - \int - \cos x \mathrm{dx}$
${F}_{2} \left(x\right) = - x \cos x + \sin x = \left[\sin x - x \cos x\right]$

$F \left(x\right) = {F}_{1} \left(x\right) - {F}_{2} = {\left[\ln \tan \left(\frac{x}{2}\right) - \sin x + x \cos x\right]}_{\frac{\pi}{6}}^{\frac{5 \pi}{8}}$