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

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

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integral of-1/9 cos(3x)

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

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

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

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

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

Substitute back

u = 3 x

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

Simplify

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

= - \frac{1}{27} sin (3 x)

Add a constant to the solution

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

\frac{\partial}{\partial x} (tan (2 x) sin (2 x))

f (x) = 4 x^{2} - 5 x

\frac{\partial}{\partial z} (cos (x z))

n = 1 \sum \infty \frac{n}{2 n ^{3} + 1}

\frac{2}{x}, at x = \frac{1}{9}

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