What is the integral of 1/3 sin(3t) ?

The integral of 1/3 sin(3t) is -1/9 cos(3t)+C

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integral of 1/3 sin(3t)

\int \frac{1}{3} sin (3 t) d t

- \frac{1}{9} cos (3 t) + C

Solution steps

\int \frac{1}{3} sin (3 t) d t

Take the constant out:

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

= \frac{1}{3} \cdot \int sin (3 t) d t

Apply u-substitution

: \int sin (u) \frac{1}{3} d u

= \frac{1}{3} \cdot \int sin (u) \frac{1}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \frac{1}{3} \cdot \int sin (u) d u

Use the common integral:

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

= \frac{1}{3} \cdot \frac{1}{3} (- cos (u))

Substitute back

u = 3 t

= \frac{1}{3} \cdot \frac{1}{3} (- cos (3 t))

Simplify

\frac{1}{3} \cdot \frac{1}{3} (- cos (3 t)) : - \frac{1}{9} cos (3 t)

= - \frac{1}{9} cos (3 t)

Add a constant to the solution

= - \frac{1}{9} cos (3 t) + C

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What is the integral of 1/3 sin(3t) ?
The integral of 1/3 sin(3t) is -1/9 cos(3t)+C