What is the integral of-1/9 sin(3x) ?

The integral of-1/9 sin(3x) is 1/27 cos(3x)+C

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integral of-1/9 sin(3x)

\int - \frac{1}{9} sin (3 x) d x

\frac{1}{27} cos (3 x) + C

Solution steps

\int - \frac{1}{9} sin (3 x) d x

Take the constant out:

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

= - \frac{1}{9} \cdot \int sin (3 x) d x

Apply u-substitution

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

= - \frac{1}{9} \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}{9} \cdot \frac{1}{3} \cdot \int sin (u) d u

Use the common integral:

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

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

Substitute back

u = 3 x

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

Simplify

- \frac{1}{9} \cdot \frac{1}{3} (- cos (3 x)) : \frac{1}{27} cos (3 x)

= \frac{1}{27} cos (3 x)

Add a constant to the solution

= \frac{1}{27} cos (3 x) + C

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What is the integral of-1/9 sin(3x) ?
The integral of-1/9 sin(3x) is 1/27 cos(3x)+C