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

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

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

\int_{0}^{\frac{π}{12}} cos (6 x) d x

\frac{1}{6}

Decimal Notation

0.16666 \dots

Solution steps

\int_{0}^{\frac{π}{12}} cos (6 x) d x

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}

\int_{0}^{2} x 3^{x} d x

\int \frac{( x ^{2} + 2 )}{x + 2} d x

\frac{\partial}{\partial x} \frac{( x - y )}{ln ( \frac{x}{y} )}

2 \frac{d ^{2} y}{d x ^{2}} - 20 \frac{d y}{d x} + 48 y = 3 sin (5 x)

\frac{s}{s ^{2} + 2 s + 2}

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