What is the integral from 0 to 1 of x^2arccos(x) ?

The integral from 0 to 1 of x^2arccos(x) is 2/9

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integral from 0 to 1 of x^2arccos(x)

\int_{0}^{1} x^{2} arccos (x) d x

\frac{2}{9}

Decimal Notation

0.22222 \dots

Solution steps

\int_{0}^{1} x^{2} arccos (x) d x

Apply Integration By Parts

: [\frac{1}{3} x^{3} arccos (x) - \int - \frac{x ^{3}}{3 1 - x ^{2}} d x]_{0}^{1}

= [\frac{1}{3} x^{3} arccos (x) - \int - \frac{x ^{3}}{3 1 - x ^{2}} d x]_{0}^{1}

\int - \frac{x ^{3}}{3 1 - x ^{2}} d x = - \frac{1}{3} (\frac{1}{3} (1 - x^{2})^{\frac{3}{2}} - 1 - x^{2})

= [\frac{1}{3} x^{3} arccos (x) - (- \frac{1}{3} (\frac{1}{3} (1 - x^{2})^{\frac{3}{2}} - 1 - x^{2}))]_{0}^{1}

Simplify

= [\frac{1}{3} x^{3} arccos (x) + \frac{1}{3} (\frac{1}{3} (1 - x^{2})^{\frac{3}{2}} - 1 - x^{2})]_{0}^{1}

Compute the boundaries:

\frac{2}{9}

= \frac{2}{9}

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What is the integral from 0 to 1 of x^2arccos(x) ?
The integral from 0 to 1 of x^2arccos(x) is 2/9