Evaluate the integral of x2 arccos x

Evaluate the following integral:

$\int x^2 \arccos x \, dx.$

First, we integrate by parts, with

$\begin{align*} u &= x \arccos x & du &= \arccos x + \frac{-x}{\sqrt{1-x^2}} \, dx \\ dv &= x \, dx & v &= \frac{x^2}{2}. \end{align*}$

Therefore, we have

$\begin{align*} &&\int x^2 \arccos x \, dx &= \frac{x^3}{2} \arccos x - \frac{1}{2} \int x^2 \arccos x \, dx + \frac{1}{2} \int \frac{x^3}{\sqrt{1-x^2}} \, dx \\ \implies && \frac{3}{2} \int x^2 \arccos x \, dx &= \frac{x^3}{2} \arccos x + \frac{1}{2} \int \frac{x^3}{\sqrt{1-x^2}} \, dx \\ \implies && \int x^2 \arccos x \, dx &= \frac{x^3}{3} \arccos x + \frac{1}{3} \int \frac{x^3}{\sqrt{1-x^2}} \, dx. \end{align*}$

Now, to evaluate the integral $\int \frac{x^3}{\sqrt{1-x^2}} \, dx$ we make a substitution. Let $s = 1-x^2$ which implies $ds = -2x \, dx$ . Therefore,

$\begin{align*} \int \frac{x^3}{\sqrt{1-x^2}} \, dx &= -\frac{1}{2} \int \frac{1-s}{\sqrt{s}} \, ds \\ &= -\frac{1}{2} \int \frac{1}{\sqrt{s}} \, ds + \frac{1}{2} \int \sqrt{s} \, ds \\ &= -\sqrt{s} + \frac{1}{3} s^{\frac{3}{2}} + C \\ &= -\sqrt{1-x^2} + \frac{1}{3} (1-x^2)^{\frac{3}{2}} + C. \end{align*}$

Plugging this back in we then have

$\begin{align*} \int x^2 \arccos x \, dx &= \frac{x^3}{3} \arccos x + \frac{1}{3} \int \frac{x^3}{\sqrt{1-x^2}} \, dx \\ &= \frac{x^3}{3} \arccos x + \frac{1}{3} \left( -\sqrt{1-x^2} + \frac{1}{3} (1-x^2)^{\frac{3}{2}} \right) + C \\ &= \frac{x^3}{3} \arccos x - \frac{1}{3} \sqrt{1-x^2} + \frac{1}{9} (1-x^2)^{\frac{3}{2}} + C \\ &= \frac{x^3}{3} \arccos x - \frac{1}{9}\sqrt{1-x^2}\left( 3 - (1-x^2) \right) + C \\ &= \frac{x^3}{3} \arccos x - \frac{2+x^2}{9} \sqrt{1-x^2} + C. \end{align*}$

Stumbling Robot

Evaluate the integral of x² arccos x

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