What is the integral from 0 to 1 of xarcsin(x^2) ?

The integral from 0 to 1 of xarcsin(x^2) is pi/4-1/2

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integral from 0 to 1 of xarcsin(x^2)

\int_{0}^{1} x arcsin (x^{2}) d x

\frac{π}{4} - \frac{1}{2}

Decimal Notation

0.28539 \dots

Solution steps

\int_{0}^{1} x arcsin (x^{2}) d x

Apply Integration By Parts

: [\frac{1}{2} x^{2} arcsin (x^{2}) - \int \frac{x ^{3}}{1 - x ^{4}} d x]_{0}^{1}

= [\frac{1}{2} x^{2} arcsin (x^{2}) - \int \frac{x ^{3}}{1 - x ^{4}} d x]_{0}^{1}

\int \frac{x ^{3}}{1 - x ^{4}} d x = - \frac{1}{2} 1 - x^{4}

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

Simplify

[\frac{1}{2} x^{2} arcsin (x^{2}) - (- \frac{1}{2} 1 - x^{4})]_{0}^{1} : [\frac{1}{2} (x^{2} arcsin (x^{2}) + 1 - x^{4})]_{0}^{1}

= [\frac{1}{2} (x^{2} arcsin (x^{2}) + 1 - x^{4})]_{0}^{1}

Compute the boundaries:

\frac{π}{4} - \frac{1}{2}

= \frac{π}{4} - \frac{1}{2}

\frac{d}{d x} (1200 x^{2} - 0.1 x)

\frac{d}{d x} (\frac{1}{x ^{3} + 1})

\frac{d y}{d x} = x y + 4 x + 5 y + 20

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

\int_{0}^{2} \frac{3}{16 - x ^{2}} d x

What is the integral from 0 to 1 of xarcsin(x^2) ?
The integral from 0 to 1 of xarcsin(x^2) is pi/4-1/2