Evaluate the integral. $$ \int \sin (9 x-2) d x $$

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Let $$ u = 9x - 2 $$. Then,  
$$
\frac{du}{dx} = 9 \quad \Longrightarrow \quad dx = \frac{du}{9}.
$$

Substitute into the integral:
$$
\int \sin(9x-2) \, dx = \int \sin(u) \cdot \frac{du}{9} = \frac{1}{9} \int \sin(u) \, du.
$$

Now, compute the integral:
$$
\int \sin(u) \, du = -\cos(u) + C.
$$

Thus,
$$
\frac{1}{9} \int \sin(u) \, du = \frac{1}{9} \left( -\cos(u) \right) + C = -\frac{1}{9}\cos(u) + C.
$$

Replace $$ u $$ with $$ 9x - 2 $$ to obtain:
$$
-\frac{1}{9}\cos(9x-2) + C.
$$
The integral of $$ \sin(9x - 2) $$ with respect to $$ x $$ is $$ -\frac{1}{9}\cos(9x - 2) + C $$, where $$ C $$ is the constant of integration.

Evaluate the integral.

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