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

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

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

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

\frac{π ^{2}}{4} - 2

Decimal Notation

0.46740 \dots

Solution steps

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

Apply Integration By Parts

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

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

\int \frac{2 x arcsin ( x )}{1 - x ^{2}} d x = 2 (- arcsin (x) 1 - x^{2} + x)

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

Compute the boundaries:

\frac{π ^{2}}{4} - 2

= \frac{π ^{2}}{4} - 2

\int \frac{1}{x ^{2} x ^{2} - 16} d x

x \to 0 lim (sin (\frac{π}{x}))

x + 2, at x = 4

\int (6 x^{2} - 5 x + 4) d x

\frac{d}{d t} (x e^{\frac{y}{z ( t )}})

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