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

The integral from 0 to r of xarcsin(x/(2r)) is ((-pi+3sqrt(3))r^2)/(12)

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

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

\frac{( - π + 3 3 ) r ^{2}}{12}

Solution steps

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

Apply u-substitution

: \int_{0}^{\frac{1}{2}} 4 r^{2} u arcsin (u) d u

= \int_{0}^{\frac{1}{2}} 4 r^{2} u arcsin (u) d u

Take the constant out:

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

= 4 r^{2} \cdot \int_{0}^{\frac{1}{2}} u arcsin (u) d u

Apply Integration By Parts

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

= 4 r^{2} [\frac{1}{2} (u^{2} arcsin (u) - 2 \cdot \int \frac{u ^{2}}{2 1 - u ^{2}} d u)]_{0}^{\frac{1}{2}}

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

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

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

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

Compute the boundaries:

\frac{- π + 3 3}{48}

= 4 r^{2} \frac{- π + 3 3}{48}

Simplify

= \frac{( - π + 3 3 ) r ^{2}}{12}

\int \frac{sec ( x ) + cos ( x )}{8 cos ( x )} d x

\frac{\partial}{\partial x} (ln (x^{2} + 1))

x \to 1 + lim (arctan (ln (x)))

\frac{9 x}{3 x + 7}

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

What is the integral from 0 to r of xarcsin(x/(2r)) ?
The integral from 0 to r of xarcsin(x/(2r)) is ((-pi+3sqrt(3))r^2)/(12)