What is the integral of (-cos(3x))/3 ?

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

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integral of (-cos(3x))/3

\int \frac{- cos ( 3 x )}{3} d x

- \frac{1}{9} sin (3 x) + C

Solution steps

\int \frac{- cos ( 3 x )}{3} d x

Take the constant out:

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

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

Simplify

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

Apply u-substitution

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

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

Use the common integral:

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

= (- \frac{1}{3}) \frac{1}{3} sin (u)

Substitute back

u = 3 x

= (- \frac{1}{3}) \frac{1}{3} sin (3 x)

Simplify

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

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

Add a constant to the solution

= - \frac{1}{9} sin (3 x) + C

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