What is the integral from 0 to 1/7 of xarcsin(7x) ?

The integral from 0 to 1/7 of xarcsin(7x) is pi/(392)

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integral from 0 to 1/7 of xarcsin(7x)

\int_{0}^{\frac{1}{7}} x arcsin (7 x) d x

\frac{π}{392}

Decimal Notation

0.00801 \dots

Solution steps

\int_{0}^{\frac{1}{7}} x arcsin (7 x) d x

Apply u-substitution

: \int_{0}^{1} \frac{u arcsin ( u )}{49} d u

= \int_{0}^{1} \frac{u arcsin ( u )}{49} d u

Take the constant out:

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

= \frac{1}{49} \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}{49} [\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}{49} [\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}{49} \cdot \frac{π}{8}

Simplify

= \frac{π}{392}

\frac{2 d y}{d x} - 4 x y = 8 x, y (0) = 15

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

\int \frac{1}{t ^{2} - 10 t + 29} d t

\frac{d ^{2} y}{d x ^{2}} = (\frac{d y}{d x})^{2}

4 y^{^{''}} + 7 y = 0

What is the integral from 0 to 1/7 of xarcsin(7x) ?
The integral from 0 to 1/7 of xarcsin(7x) is pi/(392)