# What is the antiderivative of (arccos(x/6))^2?

Aug 6, 2018

$I = x {\cos}^{-} 1 {\left(\frac{x}{6}\right)}^{2} - 12 {\cos}^{-} 1 \left(\frac{x}{6}\right) \sqrt{1 - {x}^{2} / 36} - 2 x + C$, $C \in \mathbb{R}$

#### Explanation:

$I = \int {\cos}^{- 1} {\left(\frac{x}{6}\right)}^{2} \mathrm{dx}$

Let $X = \frac{x}{6}$

$\mathrm{dx} = 6 \mathrm{dX}$

So:

$I = 6 \int {\cos}^{- 1} {\left(X\right)}^{2} \mathrm{dX}$

Using intregration by parts :
$f ' \left(X\right) = 1$, $f \left(X\right) = X$,

$g \left(X\right) = {\cos}^{-} 1 {\left(X\right)}^{2}$, $g ' \left(X\right) = \frac{- 2 {\cos}^{-} 1 \left(X\right)}{\sqrt{1 - {X}^{2}}}$

So :

$I = 6 X {\cos}^{- 1} {\left(X\right)}^{2} + 12 \int \frac{X {\cos}^{- 1} \left(X\right)}{\sqrt{1 - {X}^{2}}} \mathrm{dX}$

We will use another integration by parts,

$g \left(X\right) = {\cos}^{-} 1 \left(X\right)$, $g ' \left(X\right) = - \frac{1}{\sqrt{1 - {X}^{2}}}$

$f ' \left(X\right) = \frac{X}{\sqrt{1 - {X}^{2}}}$, $f \left(X\right) = - \sqrt{1 - {X}^{2}}$ (Here's a proof.)

So :

$I = 6 X {\cos}^{- 1} {\left(X\right)}^{2} + 12 \left(- {\cos}^{-} 1 \left(X\right) \sqrt{1 - {X}^{2}} - \int \frac{- \sqrt{1 - {X}^{2}}}{- \sqrt{1 - {X}^{2}}} \mathrm{dX}\right)$

$= 6 X {\cos}^{- 1} {\left(X\right)}^{2} - 12 {\cos}^{-} 1 \left(X\right) \sqrt{1 - {X}^{2}} - 12 \int 1 \mathrm{dX}$

$= 6 X {\cos}^{- 1} {\left(X\right)}^{2} - 12 {\cos}^{-} 1 \left(X\right) \sqrt{1 - {X}^{2}} - 12 X + C$, $C \in \mathbb{R}$

Substitute back, and finally we have :

$I = x {\cos}^{-} 1 {\left(\frac{x}{6}\right)}^{2} - 12 {\cos}^{-} 1 \left(\frac{x}{6}\right) \sqrt{1 - {x}^{2} / 36} - 2 x + C$, $C \in \mathbb{R}$