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

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

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

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

\frac{π}{4} - \frac{1}{2}

Decimal

0.28539 \dots

Solution steps

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

Apply Integration By Parts

= [y arcsin (2 y) - \int \frac{2 y}{1 - 4 y ^{2}} d y]_{0}^{\frac{1}{2}}

\int \frac{2 y}{1 - 4 y ^{2}} d y = - \frac{1}{2} 1 - 4 y^{2}

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

Simplify

[y arcsin (2 y) - (- \frac{1}{2} 1 - 4 y^{2})]_{0}^{\frac{1}{2}} : [\frac{1}{2} (2 y arcsin (2 y) + 1 - 4 y^{2})]_{0}^{\frac{1}{2}}

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

Compute the boundaries:

\frac{π}{4} - \frac{1}{2}

= \frac{π}{4} - \frac{1}{2}

What is the integral from 0 to 1/2 of arcsin(2y) ?
The integral from 0 to 1/2 of arcsin(2y) is pi/4-1/2