What is the integral of 7/9 sin(3x+6) ?

The integral of 7/9 sin(3x+6) is -7/27 cos(3x+6)+C

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integral of 7/9 sin(3x+6)

\int \frac{7}{9} sin (3 x + 6) d x

- \frac{7}{27} cos (3 x + 6) + C

Solution steps

\int \frac{7}{9} sin (3 x + 6) d x

Take the constant out:

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

= \frac{7}{9} \cdot \int sin (3 x + 6) d x

Apply u-substitution

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

= \frac{7}{9} \cdot \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{7}{9} \cdot \frac{1}{3} \cdot \int sin (u) d u

Use the common integral:

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

= \frac{7}{9} \cdot \frac{1}{3} (- cos (u))

Substitute back

u = 3 x + 6

= \frac{7}{9} \cdot \frac{1}{3} (- cos (3 x + 6))

Simplify

\frac{7}{9} \cdot \frac{1}{3} (- cos (3 x + 6)) : - \frac{7}{27} cos (3 x + 6)

= - \frac{7}{27} cos (3 x + 6)

Add a constant to the solution

= - \frac{7}{27} cos (3 x + 6) + C

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What is the integral of 7/9 sin(3x+6) ?
The integral of 7/9 sin(3x+6) is -7/27 cos(3x+6)+C