Integral of Cosine function

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$\int \cos\left(18x-3\right)\; dx$

**Answer**

$\int \cos\left(18x-3\right)\; dx$

Substitute $u=18x-3$. So, $du =18\; dx$

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

And, $\int \cos\left(18x-3\right)\; dx = \frac{1}{18}\int \cos u \;du$

$=\frac{1}{18}\sin u + C$

Substituting back $u=18x-3$

$=\frac{1}{18}\sin \left(18x-3\right)+C$