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

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

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

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

\frac{π ^{2}}{4} - 2

Decimal Notation

0.46740 \dots

Solution steps

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

Apply Integration By Parts

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

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

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

= [t arcsin^{2} (t) - 2 (- arcsin (t) 1 - t^{2} + t)]_{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(t))^2 ?
The integral from 0 to 1 of (arcsin(t))^2 is (pi^2)/4-2