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

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

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

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

2 (\frac{π}{8} - \frac{1}{4})

Decimal Notation

0.28539 \dots

Solution steps

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

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 x) d x

Apply Integration By Parts

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

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

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

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

Simplify

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

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

Compute the boundaries:

\frac{π}{8} - \frac{1}{4}

= 2 (\frac{π}{8} - \frac{1}{4})

\int \frac{v ( x )}{cos ( x )}

x \to 2 lim (x^{5} - 32 x)

2 (x^{2} + y^{2}) d x + (7 x y) d y = 0

\int_{0}^{2} \frac{x ^{2}}{x ^{2} + 4} d x

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

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