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

The integral from 0 to 1/4 of arcsin(2x) is (pi+6sqrt(3))/(24)-1/2

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integral from 0 to 1/4 of arcsin(2x)

\int_{0}^{\frac{1}{4}} arcsin (2 x) d x

\frac{π + 6 3}{24} - \frac{1}{2}

Decimal

0.06391 \dots

Solution steps

\int_{0}^{\frac{1}{4}} 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}{4}}

\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}{4}}

Simplify

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

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

Compute the boundaries:

\frac{π + 6 3}{24} - \frac{1}{2}

= \frac{π + 6 3}{24} - \frac{1}{2}

What is the integral from 0 to 1/4 of arcsin(2x) ?
The integral from 0 to 1/4 of arcsin(2x) is (pi+6sqrt(3))/(24)-1/2