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

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

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integral of-9sin(3x)

\int - 9 sin (3 x) d x

3 cos (3 x) + C

Solution steps

\int - 9 sin (3 x) d x

Take the constant out:

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

= - 9 \cdot \int sin (3 x) d x

Apply u-substitution

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

= - 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

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

Use the common integral:

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

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

Substitute back

u = 3 x

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

Simplify

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

= 3 cos (3 x)

Add a constant to the solution

= 3 cos (3 x) + C

\int 12 x ln (3 x) d x

\frac{d y}{d t} = t y^{\frac{1}{3}}

2 + e^{2 t} + e^{- 2 t}

\frac{\partial}{\partial x} (36 (\frac{x}{x + y}) - x)

(x^{2} + 1) (\frac{d y}{d x}) + 8 x (y - 1) = 0, y (0) = 6