Integral of Cosine function

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$\int \cos\left(10x+8\right)\; dx$

**Answer**

$\int \cos\left(10x+8\right)\; dx$

Substitute $u=10x+8$. So, $du =10\; dx$

Or, $\frac{1}{10}\; du = dx$

And, $\int \cos\left(10x+8\right)\; dx = \frac{1}{10}\int \cos u \;du$

$=\frac{1}{10}\sin u + C$

Substituting back $u=10x+8$

$=\frac{1}{10}\sin \left(10x+8\right)+C$