What is the integral from 0 to 1/3 of xarcsin(3x) ?

The integral from 0 to 1/3 of xarcsin(3x) is pi/(72)

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integral from 0 to 1/3 of xarcsin(3x)

\int_{0}^{\frac{1}{3}} x arcsin (3 x) d x

\frac{π}{72}

Decimal Notation

0.04363 \dots

Solution steps

\int_{0}^{\frac{1}{3}} x arcsin (3 x) d x

Apply u-substitution

: \int_{0}^{1} \frac{u arcsin ( u )}{9} d u

= \int_{0}^{1} \frac{u arcsin ( u )}{9} d u

Take the constant out:

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

= \frac{1}{9} \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}{9} [\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}{9} [\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}{9} \cdot \frac{π}{8}

Simplify

= \frac{π}{72}

\frac{e ^{x}}{9 x ^{2}}

\frac{d}{d x} (cot (7 x - 3))

x \to - 6 lim (\frac{x ^{2} - 9}{6 - x})

\int (x + 4)^{9} d x

f (t) = e^{t} \cdot sin (t)

What is the integral from 0 to 1/3 of xarcsin(3x) ?
The integral from 0 to 1/3 of xarcsin(3x) is pi/(72)