Integral of Sine function

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$\int \sin\left(18x+6\right)\; dx$

**Answer**

$\int \sin\left(18x+6\right)\; dx$

Substitute $u=18x+6$. So, $du =18\; dx$

Or, $\frac{1}{18}\; du = dx$

And, $\int \sin\left(18x+6\right)\; dx = \frac{1}{18}\int \sin u \;du$

$=-\frac{1}{18}\cos u + C$

Substituting back $u=18x+6$

$=-\frac{1}{18}\cos \left(18x+6\right)+C$