What is the integral from 0 to 1/2 of xarcsin(x) ?

The integral from 0 to 1/2 of xarcsin(x) is (-pi+3sqrt(3))/(48)

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integral from 0 to 1/2 of xarcsin(x)

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

\frac{- π + 3 3}{48}

Decimal

0.04280 \dots

Solution steps

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

Apply Integration By Parts

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

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

\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} arcsin (x) - 2 \cdot \frac{1}{4} (arcsin (x) - \frac{1}{2} sin (2 arcsin (x))))]_{0}^{\frac{1}{2}}

2 \cdot \frac{1}{4} (arcsin (x) - \frac{1}{2} sin (2 arcsin (x))) = \frac{1}{2} (arcsin (x) - \frac{1}{2} sin (2 arcsin (x)))

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

Compute the boundaries:

\frac{- π + 3 3}{48}

= \frac{- π + 3 3}{48}

\int_{0}^{2 π} (sin (x))^{2} d x

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

\int_{0}^{\frac{π}{2}} cos (θ) sin (θ) d θ

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

\int_{- π}^{π} x cos (n x) d x

What is the integral from 0 to 1/2 of xarcsin(x) ?
The integral from 0 to 1/2 of xarcsin(x) is (-pi+3sqrt(3))/(48)