Integral of Sine function

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$\int \sin\left(19x-7\right)\; dx$

**Answer**

$\int \sin\left(19x-7\right)\; dx$

Substitute $u=19x-7$. So, $du =19\; dx$

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

And, $\int \sin\left(19x-7\right)\; dx = \frac{1}{19}\int \sin u \;du$

$=-\frac{1}{19}\cos u + C$

Substituting back $u=19x-7$

$=-\frac{1}{19}\cos \left(19x-7\right)+C$