What is the integral from 0 to 1/5 of xarcsin(5x) ?

The integral from 0 to 1/5 of xarcsin(5x) is pi/(200)

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integral from 0 to 1/5 of xarcsin(5x)

\int_{0}^{\frac{1}{5}} x arcsin (5 x) d x

\frac{π}{200}

Decimal

0.01570 \dots

Solution steps

\int_{0}^{\frac{1}{5}} x arcsin (5 x) d x

Apply u-substitution

: \int_{0}^{1} \frac{u arcsin ( u )}{25} d u

= \int_{0}^{1} \frac{u arcsin ( u )}{25} d u

Take the constant out:

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

= \frac{1}{25} \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}{25} [\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}{25} [\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}{25} \cdot \frac{π}{8}

Simplify

= \frac{π}{200}

\int e^{2 x^{2}} d x

x \to 0 lim (\frac{tan ( 2 x )}{5 x})

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What is the integral from 0 to 1/5 of xarcsin(5x) ?
The integral from 0 to 1/5 of xarcsin(5x) is pi/(200)