What is the integral from 0 to pi/(12) of cos(6t) ?

The integral from 0 to pi/(12) of cos(6t) is 1/6

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integral from 0 to pi/(12) of cos(6t)

\int_{0}^{\frac{π}{12}} cos (6 t) d t

\frac{1}{6}

Decimal

0.16666 \dots

Solution steps

\int_{0}^{\frac{π}{12}} cos (6 t) d t

Apply u-substitution

: \int_{0}^{\frac{π}{2}} cos (u) \frac{1}{6} d u

= \int_{0}^{\frac{π}{2}} cos (u) \frac{1}{6} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int_{0}^{\frac{π}{2}} cos (u) d u

Use the common integral:

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

= \frac{1}{6} [sin (u)]_{0}^{\frac{π}{2}}

Compute the boundaries:

1

= \frac{1}{6} \cdot 1

Simplify

= \frac{1}{6}

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What is the integral from 0 to pi/(12) of cos(6t) ?
The integral from 0 to pi/(12) of cos(6t) is 1/6