What is the integral of sin(9t) ?

The integral of sin(9t) is -1/9 cos(9t)+C

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integral of sin(9t)

\int sin (9 t) d t

- \frac{1}{9} cos (9 t) + C

Solution steps

\int sin (9 t) d t

Apply u-substitution

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = 9 t

= \frac{1}{9} (- cos (9 t))

Simplify

= - \frac{1}{9} cos (9 t)

Add a constant to the solution

= - \frac{1}{9} cos (9 t) + C