What is the integral from 0 to x of v^2arcsin(v^3) ?

The integral from 0 to x of v^2arcsin(v^3) is (x^3arcsin(x^3)+sqrt(1-x^6)-1)/3

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integral from 0 to x of v^2arcsin(v^3)

\int_{0}^{x} v^{2} arcsin (v^{3}) d v

\frac{x ^{3} arcsin ( x ^{3} ) + 1 - x ^{6} - 1}{3}

Solution steps

\int_{0}^{x} v^{2} arcsin (v^{3}) d v

Apply Integration By Parts

: [\frac{1}{3} v^{3} arcsin (v^{3}) - \int \frac{v ^{5}}{1 - v ^{6}} d v]_{0}^{x}

= [\frac{1}{3} v^{3} arcsin (v^{3}) - \int \frac{v ^{5}}{1 - v ^{6}} d v]_{0}^{x}

\int \frac{v ^{5}}{1 - v ^{6}} d v = - \frac{1}{3} 1 - v^{6}

= [\frac{1}{3} v^{3} arcsin (v^{3}) - (- \frac{1}{3} 1 - v^{6})]_{0}^{x}

Simplify

[\frac{1}{3} v^{3} arcsin (v^{3}) - (- \frac{1}{3} 1 - v^{6})]_{0}^{x} : [\frac{1}{3} (v^{3} arcsin (v^{3}) + 1 - v^{6})]_{0}^{x}

= [\frac{1}{3} (v^{3} arcsin (v^{3}) + 1 - v^{6})]_{0}^{x}

Compute the boundaries:

\frac{1}{3} (x^{3} arcsin (x^{3}) + 1 - x^{6}) - \frac{1}{3}

= \frac{1}{3} (x^{3} arcsin (x^{3}) + 1 - x^{6}) - \frac{1}{3}

Simplify

= \frac{x ^{3} arcsin ( x ^{3} ) + 1 - x ^{6} - 1}{3}

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What is the integral from 0 to x of v^2arcsin(v^3) ?
The integral from 0 to x of v^2arcsin(v^3) is (x^3arcsin(x^3)+sqrt(1-x^6)-1)/3