What is the integral of sin(-8x) ?

The integral of sin(-8x) is 1/8 cos(8x)+C

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integral of sin(-8x)

\int sin (- 8 x) d x

\frac{1}{8} cos (8 x) + C

Solution steps

\int sin (- 8 x) d x

Apply u-substitution

: \int - \frac{1}{8} sin (u) d u

= \int - \frac{1}{8} sin (u) d u

Take the constant out:

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

= - \frac{1}{8} \cdot \int sin (u) d u

Use the common integral:

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

= - \frac{1}{8} (- cos (u))

Substitute back

u = - 8 x

= - \frac{1}{8} (- cos (- 8 x))

Simplify

- \frac{1}{8} (- cos (- 8 x)) : \frac{1}{8} cos (8 x)

= \frac{1}{8} cos (8 x)

Add a constant to the solution

= \frac{1}{8} cos (8 x) + C

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