What is the integral of-3sin(2x) ?

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

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integral of-3sin(2x)

\int - 3 sin (2 x) d x

\frac{3}{2} cos (2 x) + C

Solution steps

\int - 3 sin (2 x) d x

Take the constant out:

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

= - 3 \cdot \int sin (2 x) d x

Apply u-substitution

: \int sin (u) \frac{1}{2} d u

= - 3 \cdot \int sin (u) \frac{1}{2} d u

Take the constant out:

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

= - 3 \cdot \frac{1}{2} \cdot \int sin (u) d u

Use the common integral:

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

= - 3 \cdot \frac{1}{2} (- cos (u))

Substitute back

u = 2 x

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

Simplify

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

= \frac{3}{2} cos (2 x)

Add a constant to the solution

= \frac{3}{2} cos (2 x) + C

\int_{- 5}^{0} 25 - x^{2} d x

\frac{d}{d x} (\frac{4 x ^{2} + 8 x + 8}{x})

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

\int sec^{2} (\frac{x}{2}) d x

f (x) = - 2 x^{3}, at x = - 1