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

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

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

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

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

Solution steps

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

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 t) d t

Simplify

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

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

Use the common integral:

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

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

Substitute back

u = 8 t

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

Simplify

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

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

Add a constant to the solution

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

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