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

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

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integral of sin(-3x)

\int sin (- 3 x) d x

\frac{1}{3} cos (3 x) + C

Solution steps

\int sin (- 3 x) d x

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = - 3 x

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

Simplify

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

= \frac{1}{3} cos (3 x)

Add a constant to the solution

= \frac{1}{3} cos (3 x) + C

t \to π lim (- 9 sin (- \frac{t}{2}))

\int - 3 x + 4 x d x

x^{- \frac{3}{4}}

\int_{0}^{\frac{π}{6}} (9 sec^{2} (x)) d x

\int \frac{1}{7 - 3 x} d x