What is the integral of 8sin(6t) ?

The integral of 8sin(6t) is -4/3 cos(6t)+C

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integral of 8sin(6t)

\int 8 sin (6 t) d t

- \frac{4}{3} cos (6 t) + C

Solution steps

\int 8 sin (6 t) d t

Take the constant out:

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

= 8 \cdot \int sin (6 t) d t

Apply u-substitution

: \int sin (u) \frac{1}{6} d u

= 8 \cdot \int sin (u) \frac{1}{6} d u

Take the constant out:

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

= 8 \cdot \frac{1}{6} \cdot \int sin (u) d u

Use the common integral:

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

= 8 \cdot \frac{1}{6} (- cos (u))

Substitute back

u = 6 t

= 8 \cdot \frac{1}{6} (- cos (6 t))

Simplify

8 \cdot \frac{1}{6} (- cos (6 t)) : - \frac{4}{3} cos (6 t)

= - \frac{4}{3} cos (6 t)

Add a constant to the solution

= - \frac{4}{3} cos (6 t) + C

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