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

= - 9 \cdot \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

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

Use the common integral:

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

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

Substitute back

u = - 3 x

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

Simplify

- 9 (- \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

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