What is the integral from-1 to 1 of xarccos(x) ?

The integral from-1 to 1 of xarccos(x) is -pi/4

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integral from-1 to 1 of xarccos(x)

\int_{- 1}^{1} x arccos (x) d x

- \frac{π}{4}

Decimal

- 0.78539 \dots

Solution steps

\int_{- 1}^{1} x arccos (x) d x

Apply Integration By Parts

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

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

\int - \frac{x ^{2}}{2 1 - x ^{2}} d x = - \frac{1}{4} (arcsin (x) - \frac{1}{2} sin (2 arcsin (x)))

= [\frac{1}{2} x^{2} arccos (x) - (- \frac{1}{4} (arcsin (x) - \frac{1}{2} sin (2 arcsin (x))))]_{- 1}^{1}

Simplify

= [\frac{1}{2} x^{2} arccos (x) + \frac{1}{4} (arcsin (x) - \frac{1}{2} sin (2 arcsin (x)))]_{- 1}^{1}

Compute the boundaries:

- \frac{π}{4}

= - \frac{π}{4}

\int_{0}^{1} \frac{x}{( 1 - x )} d x

\int_{0}^{2} x^{2} - 40 d x

\int_{1}^{3} x^{2} - 3 x + 1 d x

\int_{3}^{4} (3 x^{2} - 1) d x

\int_{0}^{\frac{π}{2}} 3 cos^{3} (x) d x

What is the integral from-1 to 1 of xarccos(x) ?
The integral from-1 to 1 of xarccos(x) is -pi/4