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

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

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integral from 0 to 1 of x*arcsin(x^2)

\int_{0}^{1} x \cdot 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}

x \to + 3 lim (π)

\int \frac{6 x ^{3} + 3 x ^{2} - 47 x - 18}{x ^{2} - 9} d x

x \to 1 lim (\frac{arctan ( x ) - \frac{π}{4}}{x - 1})

\int_{- r}^{r} π (r^{2} - x^{2}) d x

f (x) = 7 e^{x^{2} - 5 x + 8}, at x = 0

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