What is the integral from 0 to 1/9 of arccos(9x) ?

The integral from 0 to 1/9 of arccos(9x) is 1/9

Popular >

integral from 0 to 1/9 of arccos(9x)

\int_{0}^{\frac{1}{9}} arccos (9 x) d x

\frac{1}{9}

Decimal

0.11111 \dots

Solution steps

\int_{0}^{\frac{1}{9}} arccos (9 x) d x

Apply Integration By Parts

: [x arccos (9 x) - \int - \frac{9 x}{1 - 81 x ^{2}} d x]_{0}^{\frac{1}{9}}

= [x arccos (9 x) - \int - \frac{9 x}{1 - 81 x ^{2}} d x]_{0}^{\frac{1}{9}}

\int - \frac{9 x}{1 - 81 x ^{2}} d x = \frac{1}{9} 1 - 81 x^{2}

= [x arccos (9 x) - \frac{1}{9} 1 - 81 x^{2}]_{0}^{\frac{1}{9}}

Simplify

[x arccos (9 x) - \frac{1}{9} 1 - 81 x^{2}]_{0}^{\frac{1}{9}} : [\frac{1}{9} (9 x arccos (9 x) - 1 - 81 x^{2})]_{0}^{\frac{1}{9}}

= [\frac{1}{9} (9 x arccos (9 x) - 1 - 81 x^{2})]_{0}^{\frac{1}{9}}

Compute the boundaries:

\frac{1}{9}

= \frac{1}{9}

\int_{0}^{2 π} 5 t 2 d t

\int_{0}^{2} \frac{2 x ^{2} - 6 x}{2} d x

\int_{0}^{t} (t - u)^{2} cos (2 u) d u

\int_{- 8}^{8} \frac{1}{x} d x

\int_{- 3}^{- 2} \frac{y + 2}{y ^{2} + 4 y} d y

What is the integral from 0 to 1/9 of arccos(9x) ?
The integral from 0 to 1/9 of arccos(9x) is 1/9