What is the integral of-6sin(3x) ?

The integral of-6sin(3x) is 2cos(3x)+C

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integral of-6sin(3x)

\int - 6 sin (3 x) d x

2 cos (3 x) + C

Solution steps

\int - 6 sin (3 x) d x

Take the constant out:

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

= - 6 \cdot \int sin (3 x) d x

Apply u-substitution

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

= - 6 \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

= - 6 \cdot \frac{1}{3} \cdot \int sin (u) d u

Use the common integral:

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

= - 6 \cdot \frac{1}{3} (- cos (u))

Substitute back

u = 3 x

= - 6 \cdot \frac{1}{3} (- cos (3 x))

Simplify

- 6 \cdot \frac{1}{3} (- cos (3 x)) : 2 cos (3 x)

= 2 cos (3 x)

Add a constant to the solution

= 2 cos (3 x) + C

\frac{d}{d x} (e^{2 x} x)

(2, - 4), (6, 0)

\frac{\partial}{\partial x} (2 x cos (x^{2} + y^{2}))

x \to - 3 + lim (\frac{x ^{2} - 9}{x + 3})

x y + y^{^{'}} = 142 x