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

The integral from 0 to 1/4 of 2arcsin(4y) is 2(pi/8-1/4)

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

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

2 (\frac{π}{8} - \frac{1}{4})

Decimal Notation

0.28539 \dots

Solution steps

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

Apply Integration By Parts

: [y arcsin (4 y) - \int \frac{4 y}{1 - 16 y ^{2}} d y]_{0}^{\frac{1}{4}}

= 2 [y arcsin (4 y) - \int \frac{4 y}{1 - 16 y ^{2}} d y]_{0}^{\frac{1}{4}}

\int \frac{4 y}{1 - 16 y ^{2}} d y = - \frac{1}{4} 1 - 16 y^{2}

= 2 [y arcsin (4 y) - (- \frac{1}{4} 1 - 16 y^{2})]_{0}^{\frac{1}{4}}

Simplify

2 [y arcsin (4 y) - (- \frac{1}{4} 1 - 16 y^{2})]_{0}^{\frac{1}{4}} : 2 [\frac{1}{4} (4 y arcsin (4 y) + 1 - 16 y^{2})]_{0}^{\frac{1}{4}}

= 2 [\frac{1}{4} (4 y arcsin (4 y) + 1 - 16 y^{2})]_{0}^{\frac{1}{4}}

Compute the boundaries:

\frac{π}{8} - \frac{1}{4}

= 2 (\frac{π}{8} - \frac{1}{4})

\int sin (\frac{π x ^{2}}{2}) d x

f (x) = \frac{x + 3}{x - 3}

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

arctan (x^{4})

y = \frac{1}{x + 1} + 1

What is the integral from 0 to 1/4 of 2arcsin(4y) ?
The integral from 0 to 1/4 of 2arcsin(4y) is 2(pi/8-1/4)