What is the integral from 0 to 1/8 of arccos(8x) ?

The integral from 0 to 1/8 of arccos(8x) is 1/8

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integral from 0 to 1/8 of arccos(8x)

\int_{0}^{\frac{1}{8}} arccos (8 x) d x

\frac{1}{8}

Decimal

0.125

Solution steps

\int_{0}^{\frac{1}{8}} arccos (8 x) d x

Apply Integration By Parts

: [x arccos (8 x) - \int - \frac{8 x}{1 - 64 x ^{2}} d x]_{0}^{\frac{1}{8}}

= [x arccos (8 x) - \int - \frac{8 x}{1 - 64 x ^{2}} d x]_{0}^{\frac{1}{8}}

\int - \frac{8 x}{1 - 64 x ^{2}} d x = \frac{1}{8} 1 - 64 x^{2}

= [x arccos (8 x) - \frac{1}{8} 1 - 64 x^{2}]_{0}^{\frac{1}{8}}

Simplify

[x arccos (8 x) - \frac{1}{8} 1 - 64 x^{2}]_{0}^{\frac{1}{8}} : [\frac{1}{8} (8 x arccos (8 x) - 1 - 64 x^{2})]_{0}^{\frac{1}{8}}

= [\frac{1}{8} (8 x arccos (8 x) - 1 - 64 x^{2})]_{0}^{\frac{1}{8}}

Compute the boundaries:

\frac{1}{8}

= \frac{1}{8}

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What is the integral from 0 to 1/8 of arccos(8x) ?
The integral from 0 to 1/8 of arccos(8x) is 1/8