What is the integral of 6sin(3(x-2)) ?

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

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integral of 6sin(3(x-2))

\int 6 sin (3 (x - 2)) d x

- 2 cos (3 x - 6) + C

Solution steps

\int 6 sin (3 (x - 2)) d x

Take the constant out:

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

= 6 \cdot \int sin (3 (x - 2)) d x

Expand

3 (x - 2) : 3 x - 6

= 6 \cdot \int sin (3 x - 6) d x

Apply u-substitution

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

= 6 \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

= 6 \cdot \frac{1}{3} \cdot \int sin (u) d u

Use the common integral:

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

= 6 \cdot \frac{1}{3} (- cos (u))

Substitute back

u = 3 x - 6

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

Simplify

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

= - 2 cos (3 x - 6)

Add a constant to the solution

= - 2 cos (3 x - 6) + C

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What is the integral of 6sin(3(x-2)) ?
The integral of 6sin(3(x-2)) is -2cos(3x-6)+C