What is the integral of sin(3y) ?

The integral of sin(3y) is -1/3 cos(3y)+C

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integral of sin(3y)

\int sin (3 y) d y

- \frac{1}{3} cos (3 y) + C

Solution steps

\int sin (3 y) d y

Apply u-substitution

= \int sin (u) \frac{1}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int sin (u) d u

Use the common integral:

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

= \frac{1}{3} (- cos (u))

Substitute back

u = 3 y

= \frac{1}{3} (- cos (3 y))

Simplify

= - \frac{1}{3} cos (3 y)

Add a constant to the solution

= - \frac{1}{3} cos (3 y) + C