Integral of Sine function

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$\int \sin\left(10x+3\right)\; dx$

**Answer**

$\int \sin\left(10x+3\right)\; dx$

Substitute $u=10x+3$. So, $du =10\; dx$

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

And, $\int \sin\left(10x+3\right)\; dx = \frac{1}{10}\int \sin u \;du$

$=-\frac{1}{10}\cos u + C$

Substituting back $u=10x+3$

$=-\frac{1}{10}\cos \left(10x+3\right)+C$