What is the integral from 0 to pi/(24) of cos(6x) ?

The integral from 0 to pi/(24) of cos(6x) is 1/(6sqrt(2))

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integral from 0 to pi/(24) of cos(6x)

\int_{0}^{\frac{π}{24}} cos (6 x) d x

\frac{1}{6 2}

Solution steps

\int_{0}^{\frac{π}{24}} cos (6 x) d x

Apply u-substitution

: \int_{0}^{\frac{π}{4}} cos (u) \frac{1}{6} d u

= \int_{0}^{\frac{π}{4}} 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}{4}} cos (u) d u

Use the common integral:

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

= \frac{1}{6} [sin (u)]_{0}^{\frac{π}{4}}

Compute the boundaries:

\frac{1}{2}

= \frac{1}{6} \cdot \frac{1}{2}

Simplify

= \frac{1}{6 2}

\frac{\partial}{\partial z} (3 y e^{- z})

\frac{d y}{d x} = (\frac{( 2 y + 3 )}{4 x + 5})^{2}

\frac{d}{d x} (ln (\frac{3 x + 1}{x ^{2} - 2}))

3 x e^{x} csc (x)

x \to 0 lim (sin^{2} (x) (\frac{1}{x}))

What is the integral from 0 to pi/(24) of cos(6x) ?
The integral from 0 to pi/(24) of cos(6x) is 1/(6sqrt(2))