What is the integral from 0 to pi/(18) of cos(9x) ?

The integral from 0 to pi/(18) of cos(9x) is 1/9

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integral from 0 to pi/(18) of cos(9x)

\int_{0}^{\frac{π}{18}} cos (9 x) d x

\frac{1}{9}

Decimal Notation

0.11111 \dots

Solution steps

\int_{0}^{\frac{π}{18}} cos (9 x) d x

Apply u-substitution

: \int_{0}^{\frac{π}{2}} cos (u) \frac{1}{9} d u

= \int_{0}^{\frac{π}{2}} cos (u) \frac{1}{9} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int_{0}^{\frac{π}{2}} cos (u) d u

Use the common integral:

\int cos (u) d u = sin (u)

= \frac{1}{9} [sin (u)]_{0}^{\frac{π}{2}}

Compute the boundaries:

1

= \frac{1}{9} \cdot 1

Simplify

= \frac{1}{9}

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What is the integral from 0 to pi/(18) of cos(9x) ?
The integral from 0 to pi/(18) of cos(9x) is 1/9