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

The integral from 0 to r of xarcsin((r-x)/r) is r(-(pir)/8+(pi/2-1)r)

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

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

r (- \frac{π r}{8} + (\frac{π}{2} - 1) r)

Solution steps

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

Apply u-substitution

: \int_{1}^{0} - r (- r u + r) arcsin (u) d u

= \int_{1}^{0} - r (- r u + r) arcsin (u) d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{0}^{1} - r (- r u + r) arcsin (u) d u

Take the constant out:

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

= - (- r \cdot \int_{0}^{1} (- r u + r) arcsin (u) d u)

Expand

(- r u + r) arcsin (u) : - r u arcsin (u) + r arcsin (u)

= - (- r \cdot \int_{0}^{1} - r u arcsin (u) + r arcsin (u) d u)

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - (- r (- \int_{0}^{1} r u arcsin (u) d u + \int_{0}^{1} r arcsin (u) d u))

\int_{0}^{1} r u arcsin (u) d u = \frac{π r}{8}

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

= - (- r (- \frac{π r}{8} + r (\frac{π}{2} - 1)))

Simplify

= r (- \frac{π r}{8} + (\frac{π}{2} - 1) r)

\frac{d}{d x} (\frac{1}{sin ( 2 x )})

\frac{e ^{- s} - e ^{- 2 s}}{s ( s + 2 ) ( s + 1 )}

\int_{1}^{2} x 8^{x^{2}} d x

y^{^{'}} = \frac{y ( x - y )}{x ^{2}}

\frac{d y}{d x} = 2 + 3 y

What is the integral from 0 to r of xarcsin((r-x)/r) ?
The integral from 0 to r of xarcsin((r-x)/r) is r(-(pir)/8+(pi/2-1)r)