What is the integral from 0 to 1/8 of xarcsin(8x) ?

The integral from 0 to 1/8 of xarcsin(8x) is pi/(512)

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integral from 0 to 1/8 of xarcsin(8x)

\int_{0}^{\frac{1}{8}} x arcsin (8 x) d x

\frac{π}{512}

Decimal Notation

0.00613 \dots

Solution steps

\int_{0}^{\frac{1}{8}} x arcsin (8 x) d x

Apply u-substitution

: \int_{0}^{1} \frac{u arcsin ( u )}{64} d u

= \int_{0}^{1} \frac{u arcsin ( u )}{64} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{64} \cdot \int_{0}^{1} u arcsin (u) d u

Apply Integration By Parts

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

= \frac{1}{64} [\frac{1}{2} u^{2} arcsin (u) - \int \frac{u ^{2}}{2 1 - u ^{2}} d u]_{0}^{1}

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

= \frac{1}{64} [\frac{1}{2} u^{2} arcsin (u) - \frac{1}{4} (arcsin (u) - \frac{1}{2} sin (2 arcsin (u)))]_{0}^{1}

Compute the boundaries:

\frac{π}{8}

= \frac{1}{64} \cdot \frac{π}{8}

Simplify

= \frac{π}{512}

\frac{d}{d x} ((3 x^{2} - 9)^{- 15})

y = (x^{2} - 2) (x^{3} - 3 x), (- 2, - 4)

\frac{d}{d x} (3 x^{2} - 4 x + 2)

y = x^{2} - 3 x + 3, (4, 7)

\frac{d ^{2}}{d x ^{2}} (4 x + 1)

What is the integral from 0 to 1/8 of xarcsin(8x) ?
The integral from 0 to 1/8 of xarcsin(8x) is pi/(512)