Integral of Cosine function

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$\int \cos\left(19x-8\right)\; dx$

**Answer**

$\int \cos\left(19x-8\right)\; dx$

Substitute $u=19x-8$. So, $du =19\; dx$

Or, $\frac{1}{19}\; du = dx$

And, $\int \cos\left(19x-8\right)\; dx = \frac{1}{19}\int \cos u \;du$

$=\frac{1}{19}\sin u + C$

Substituting back $u=19x-8$

$=\frac{1}{19}\sin \left(19x-8\right)+C$