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

The integral from 0 to 1 of arcsin(sqrt(x)) is pi/4

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integral from 0 to 1 of arcsin(sqrt(x))

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

\frac{π}{4}

Solution steps

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

Apply u-substitution

: \int_{0}^{1} arcsin (u) \cdot 2 u d u

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

\int \frac{u ^{2}}{2 1 - u ^{2}} d u = \frac{1}{4} (arcsin (u) - \frac{1}{2} sin (2 arcsin (u)))

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

Compute the boundaries:

\frac{π}{8}

= 2 \cdot \frac{π}{8}

Simplify

= \frac{π}{4}

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