What is the integral from 0 to 1/7 of arccos(7x) ?

The integral from 0 to 1/7 of arccos(7x) is 1/7

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integral from 0 to 1/7 of arccos(7x)

\int_{0}^{\frac{1}{7}} arccos (7 x) d x

\frac{1}{7}

Solution steps

\int_{0}^{\frac{1}{7}} arccos (7 x) d x

Apply Integration By Parts

: [x arccos (7 x) - \int - \frac{7 x}{1 - 49 x ^{2}} d x]_{0}^{\frac{1}{7}}

= [x arccos (7 x) - \int - \frac{7 x}{1 - 49 x ^{2}} d x]_{0}^{\frac{1}{7}}

\int - \frac{7 x}{1 - 49 x ^{2}} d x = \frac{1}{7} 1 - 49 x^{2}

= [x arccos (7 x) - \frac{1}{7} 1 - 49 x^{2}]_{0}^{\frac{1}{7}}

Simplify

[x arccos (7 x) - \frac{1}{7} 1 - 49 x^{2}]_{0}^{\frac{1}{7}} : [\frac{1}{7} (7 x arccos (7 x) - 1 - 49 x^{2})]_{0}^{\frac{1}{7}}

= [\frac{1}{7} (7 x arccos (7 x) - 1 - 49 x^{2})]_{0}^{\frac{1}{7}}

Compute the boundaries:

\frac{1}{7}

= \frac{1}{7}

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What is the integral from 0 to 1/7 of arccos(7x) ?
The integral from 0 to 1/7 of arccos(7x) is 1/7