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

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

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

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

\frac{π}{2} - 1

Solution steps

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

Take the constant out:

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

= 2 \cdot \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}}

= 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}

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

Simplify

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

= 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}

= 2 (\frac{π}{4} - \frac{1}{2})

Simplify

= \frac{π}{2} - 1

\int \frac{1}{x ( 49 + x )} d x

9 e^{(- 4 x)}

\int (e^{x} x + e^{x}) sin (x) d x

\int_{0}^{1} \frac{6}{4 y - 1} d y

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