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

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

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

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

\frac{π}{2} - 1

Decimal

0.57079 \dots

Solution steps

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

Take the constant out:

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

= 4 \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}}

= 4 [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}

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

Simplify

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

= 4 [\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}

= 4 (\frac{π}{8} - \frac{1}{4})

Simplify

= \frac{π}{2} - 1

\int (sin (2) x^{2})^{2} x cos (2) x^{2} d x

\frac{d y}{d x} + y^{3} x + 3 y = 0

(t e^{t})^{^{'}}

\int 5 x^{3} (ln (x) - 2) d x

x \to \infty lim (\frac{30}{1 + x ^{10}})

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