What is the integral of 1/6 sin(6t) ?

The integral of 1/6 sin(6t) is -1/36 cos(6t)+C

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integral of 1/6 sin(6t)

\int \frac{1}{6} sin (6 t) d t

- \frac{1}{36} cos (6 t) + C

Solution steps

\int \frac{1}{6} sin (6 t) d t

Take the constant out:

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

= \frac{1}{6} \cdot \int sin (6 t) d t

Apply u-substitution

: \int sin (u) \frac{1}{6} d u

= \frac{1}{6} \cdot \int sin (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 \frac{1}{6} \cdot \int sin (u) d u

Use the common integral:

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

= \frac{1}{6} \cdot \frac{1}{6} (- cos (u))

Substitute back

u = 6 t

= \frac{1}{6} \cdot \frac{1}{6} (- cos (6 t))

Simplify

\frac{1}{6} \cdot \frac{1}{6} (- cos (6 t)) : - \frac{1}{36} cos (6 t)

= - \frac{1}{36} cos (6 t)

Add a constant to the solution

= - \frac{1}{36} cos (6 t) + C

f (x) = 3 + x^{2}

y = x^{2} - x + 2, at x = 2

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

x \to 0 lim (x sin (\frac{8}{x}))

(6 + x) \frac{d y}{d x} = 3 y

What is the integral of 1/6 sin(6t) ?
The integral of 1/6 sin(6t) is -1/36 cos(6t)+C