What is the integral from 0 to 1/4 of 5arcsin(4x) ?

The integral from 0 to 1/4 of 5arcsin(4x) is 5(pi/8-1/4)

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integral from 0 to 1/4 of 5arcsin(4x)

\int_{0}^{\frac{1}{4}} 5 arcsin (4 x) d x

5 (\frac{π}{8} - \frac{1}{4})

Decimal Notation

0.71349 \dots

Solution steps

\int_{0}^{\frac{1}{4}} 5 arcsin (4 x) d x

Take the constant out:

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

= 5 \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}}

= 5 [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}

= 5 [x arcsin (4 x) - (- \frac{1}{4} 1 - 16 x^{2})]_{0}^{\frac{1}{4}}

Simplify

5 [x arcsin (4 x) - (- \frac{1}{4} 1 - 16 x^{2})]_{0}^{\frac{1}{4}} : 5 [\frac{1}{4} (4 x arcsin (4 x) + 1 - 16 x^{2})]_{0}^{\frac{1}{4}}

= 5 [\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}

= 5 (\frac{π}{8} - \frac{1}{4})

\int \frac{6 x ^{3} + 5 x ^{2} - 47 x - 36}{x ^{2} - 9} d x

10 - x, 1

arccot (\frac{x}{6})

x \to - \infty lim (x^{4})

f (x) = 4 x^{3} + 3 x, at x = - 4, - 1

What is the integral from 0 to 1/4 of 5arcsin(4x) ?
The integral from 0 to 1/4 of 5arcsin(4x) is 5(pi/8-1/4)