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

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

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

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

\frac{π}{4} - \frac{1}{2}

Decimal Notation

0.28539 \dots

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

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

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

Compute the boundaries:

\frac{π}{4} - \frac{1}{2}

= \frac{π}{4} - \frac{1}{2}

\int \frac{4 x + 1}{( 2 x ^{2} + x ) ^{5}} d x

\frac{\partial}{\partial x} (8 x^{2} + 14 x y + 5 y^{2})

x^{2} + y^{2} = 25, (4, 3)

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

\frac{\partial}{\partial x} (5 e^{x y + 4})

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