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

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

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integral of-1/3 cos(3x)

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

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

Solution steps

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

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

Use the common integral:

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

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

Substitute back

u = 3 x

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

Simplify

- \frac{1}{3} \cdot \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

\int \frac{2 x ^{2}}{3 - 4 x ^{3}} d x

- 2

\frac{d}{d x} (arctan (x y))

\int \frac{( 3 x - 2 )}{5 x ^{2} - 3 x + 2} d x

f (x) = \frac{3}{1 + x ^{2}}

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