What is the integral of (arcsin(y))/(2sqrt(1-y^2)) ?

The integral of (arcsin(y))/(2sqrt(1-y^2)) is 1/4 arcsin^2(y)+C

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integral of (arcsin(y))/(2sqrt(1-y^2))

\int \frac{arcsin ( y )}{2 1 - y ^{2}} d y

\frac{1}{4} arcsin^{2} (y) + C

Solution steps

\int \frac{arcsin ( y )}{2 1 - y ^{2}} d y

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{arcsin ( y )}{1 - y ^{2}} d y

Apply u-substitution

: \int u d u

= \frac{1}{2} \cdot \int u d u

Apply the Power Rule

: \frac{u ^{2}}{2}

= \frac{1}{2} \cdot \frac{u ^{2}}{2}

Substitute back

u = arcsin (y)

= \frac{1}{2} \cdot \frac{arcsin ^{2} ( y )}{2}

Simplify

\frac{1}{2} \cdot \frac{arcsin ^{2} ( y )}{2} : \frac{1}{4} arcsin^{2} (y)

= \frac{1}{4} arcsin^{2} (y)

Add a constant to the solution

= \frac{1}{4} arcsin^{2} (y) + C

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What is the integral of (arcsin(y))/(2sqrt(1-y^2)) ?
The integral of (arcsin(y))/(2sqrt(1-y^2)) is 1/4 arcsin^2(y)+C