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 Notation

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}}

= [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}

9 sin (θ) cos (θ)

\frac{d}{d x} (x ln (x) - 1)

\int t^{2} e^{- 2 t} d t

\int \frac{( e ^{2 x} - 1 )}{( e ^{x} + e ^{- x} )} d x

\frac{d}{d x} (\frac{2 x}{3 x ^{2} - 12})

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