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

The integral from 0 to 1/(sqrt(2) of)2xarcsin(x^2) is (pi+6sqrt(3))/(12)-1

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

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

\frac{π + 6 3}{12} - 1

Decimal Notation

0.12782 \dots

Solution steps

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

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

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

Simplify

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

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

Compute the boundaries:

\frac{π + 6 3}{24} - \frac{1}{2}

= 2 (\frac{π + 6 3}{24} - \frac{1}{2})

Simplify

= \frac{π + 6 3}{12} - 1

\frac{d}{d x} (ln (\frac{x}{x + 1}))

\int (\frac{10}{x} + \frac{x}{10}) d x

x \to \infty lim (x^{2} 9)

\int \frac{x ^{2}}{( 6 - x ^{3} ) ^{5}} d x

y = ln (\frac{x}{x ^{2} + 1})

What is the integral from 0 to 1/(sqrt(2) of)2xarcsin(x^2) ?
The integral from 0 to 1/(sqrt(2) of)2xarcsin(x^2) is (pi+6sqrt(3))/(12)-1