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

The integral of-(sin(8x))/8 is 1/64 cos(8x)+C

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integral of-(sin(8x))/8

\int - \frac{sin ( 8 x )}{8} d x

\frac{1}{64} cos (8 x) + C

Solution steps

\int - \frac{sin ( 8 x )}{8} d x

Take the constant out:

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = 8 x

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

Simplify

- \frac{1}{8} \cdot \frac{1}{8} (- cos (8 x)) : \frac{1}{64} cos (8 x)

= \frac{1}{64} cos (8 x)

Add a constant to the solution

= \frac{1}{64} cos (8 x) + C

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What is the integral of-(sin(8x))/8 ?
The integral of-(sin(8x))/8 is 1/64 cos(8x)+C