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

The integral from 0 to 1/2 of arcsin(y) is pi/(12)+(sqrt(3))/2-1

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integral from 0 to 1/2 of arcsin(y)

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

\frac{π}{12} + \frac{3}{2} - 1

Decimal

0.12782 \dots

Solution steps

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

Apply Integration By Parts

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

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

\int \frac{y}{1 - y ^{2}} d y = - 1 - y^{2}

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

Simplify

= [y arcsin (y) + 1 - y^{2}]_{0}^{\frac{1}{2}}

Compute the boundaries:

\frac{π}{12} + \frac{3}{2} - 1

= \frac{π}{12} + \frac{3}{2} - 1

\int_{1 - 5 x}^{1} \frac{u ^{3}}{1 + u ^{2}} d u

\int_{1}^{2} 5 x 5 - x^{2} d x

\int_{1}^{5} \frac{1}{1 + x - 1} d x

\int_{- 2}^{2} 1 - x^{2} d x

\int_{0}^{l n (2)} (e^{t} - e^{- t}) d t

What is the integral from 0 to 1/2 of arcsin(y) ?
The integral from 0 to 1/2 of arcsin(y) is pi/(12)+(sqrt(3))/2-1