# How do you find the integral of int [arccos(x)]^n if is an integer?

Oct 20, 2015

${I}_{n} = x {\left(\arccos x\right)}^{n} - n {\left(\arccos x\right)}^{n - 1} \sqrt{1 - {x}^{2}} - n \left(n - 1\right) {I}_{n - 2}$

#### Explanation:

$u = {\left(\arccos x\right)}^{n} \implies \mathrm{du} = n {\left(\arccos x\right)}^{n - 1} \cdot \left(- \frac{1}{\sqrt{1 - {x}^{2}}}\right) \mathrm{dx}$

$\mathrm{dv} = \mathrm{dx} \implies v = x$

${I}_{n} = x {\left(\arccos x\right)}^{n} - \int x \cdot n {\left(\arccos x\right)}^{n - 1} \cdot \left(- \frac{1}{\sqrt{1 - {x}^{2}}}\right) \mathrm{dx}$

${I}_{n} = x {\left(\arccos x\right)}^{n} + n \int {\left(\arccos x\right)}^{n - 1} \cdot \frac{x}{\sqrt{1 - {x}^{2}}} \mathrm{dx}$

$u = {\left(\arccos x\right)}^{n - 1} \implies \mathrm{du} = \left(n - 1\right) {\left(\arccos x\right)}^{n - 2} \cdot \left(- \frac{1}{\sqrt{1 - {x}^{2}}}\right) \mathrm{dx}$

$\mathrm{dv} = \frac{x}{\sqrt{1 - {x}^{2}}} \mathrm{dx}$

$v = \int \frac{x}{\sqrt{1 - {x}^{2}}} \mathrm{dx}$

$1 - {x}^{2} = t \implies - 2 x \mathrm{dx} = \mathrm{dt} \implies x \mathrm{dx} = - \frac{\mathrm{dt}}{2}$

$v = - \frac{1}{2} \int \frac{\mathrm{dt}}{\sqrt{t}} = - \frac{1}{2} {t}^{\frac{1}{2}} / \left(\frac{1}{2}\right) = - \sqrt{t} = - \sqrt{1 - {x}^{2}}$

${I}_{n} = x {\left(\arccos x\right)}^{n} + n \left[- {\left(\arccos x\right)}^{n - 1} \sqrt{1 - {x}^{2}} - \int \left(- \sqrt{1 - {x}^{2}}\right) \left(n - 1\right) {\left(\arccos x\right)}^{n - 2} \cdot \left(- \frac{1}{\sqrt{1 - {x}^{2}}}\right) \mathrm{dx}\right]$

${I}_{n} = x {\left(\arccos x\right)}^{n} + n \left[- {\left(\arccos x\right)}^{n - 1} \sqrt{1 - {x}^{2}} - \left(n - 1\right) \int {\left(\arccos x\right)}^{n - 2} \mathrm{dx}\right]$

${I}_{n} = x {\left(\arccos x\right)}^{n} - n {\left(\arccos x\right)}^{n - 1} \sqrt{1 - {x}^{2}} - n \left(n - 1\right) {I}_{n - 2}$

We have recurrent formula, we only have to find ${I}_{1}$ and ${I}_{2}$ and that's trivial using Integration By Parts.