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

The integral from 0 to 1/2 of arcsin^2(x) is (pi^2)/(72)+pi/(2sqrt(3))-1

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

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

\frac{π ^{2}}{72} + \frac{π}{2 3} - 1

Decimal

0.04397 \dots

Solution steps

\int_{0}^{\frac{1}{2}} 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}^{\frac{1}{2}}

\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}^{\frac{1}{2}}

Compute the boundaries:

\frac{π ^{2}}{72} + \frac{π}{2 3} - 1

= \frac{π ^{2}}{72} + \frac{π}{2 3} - 1

What is the integral from 0 to 1/2 of arcsin^2(x) ?
The integral from 0 to 1/2 of arcsin^2(x) is (pi^2)/(72)+pi/(2sqrt(3))-1