Integral of Sine function

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$\int \sin\left(17x+9\right)\; dx$

**Answer**

$\int \sin\left(17x+9\right)\; dx$

Substitute $u=17x+9$. So, $du =17\; dx$

Or, $\frac{1}{17}\; du = dx$

And, $\int \sin\left(17x+9\right)\; dx = \frac{1}{17}\int \sin u \;du$

$=-\frac{1}{17}\cos u + C$

Substituting back $u=17x+9$

$=-\frac{1}{17}\cos \left(17x+9\right)+C$