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

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

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

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

\frac{π ^{2}}{4} - 2

Decimal

0.46740 \dots

Solution steps

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

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