What is the integral of 9sin(6x) ?

The integral of 9sin(6x) is -3/2 cos(6x)+C

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integral of 9sin(6x)

\int 9 sin (6 x) d x

- \frac{3}{2} cos (6 x) + C

Solution steps

\int 9 sin (6 x) d x

Take the constant out:

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

= 9 \cdot \int sin (6 x) d x

Apply u-substitution

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

= 9 \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

= 9 \cdot \frac{1}{6} \cdot \int sin (u) d u

Use the common integral:

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

= 9 \cdot \frac{1}{6} (- cos (u))

Substitute back

u = 6 x

= 9 \cdot \frac{1}{6} (- cos (6 x))

Simplify

9 \cdot \frac{1}{6} (- cos (6 x)) : - \frac{3}{2} cos (6 x)

= - \frac{3}{2} cos (6 x)

Add a constant to the solution

= - \frac{3}{2} cos (6 x) + C

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