What is the integral from 0 to 1/6 of xarcsin(6x) ?

The integral from 0 to 1/6 of xarcsin(6x) is pi/(288)

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integral from 0 to 1/6 of xarcsin(6x)

\int_{0}^{\frac{1}{6}} x arcsin (6 x) d x

\frac{π}{288}

Decimal Notation

0.01090 \dots

Solution steps

\int_{0}^{\frac{1}{6}} x arcsin (6 x) d x

Apply u-substitution

: \int_{0}^{1} \frac{u arcsin ( u )}{36} d u

= \int_{0}^{1} \frac{u arcsin ( u )}{36} d u

Take the constant out:

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

= \frac{1}{36} \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}{36} [\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}{36} [\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}{36} \cdot \frac{π}{8}

Simplify

= \frac{π}{288}

\int (cos (x) - sin (2 x)) d x

\frac{d y}{d x} + 2 y = 35 sin (x) + 15 cos (x), y (0) = 9

\frac{\partial}{\partial y} (e^{- x} sec (y))

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

\frac{d}{d x} (sin^{2} (2 x))

What is the integral from 0 to 1/6 of xarcsin(6x) ?
The integral from 0 to 1/6 of xarcsin(6x) is pi/(288)