Integral of $$$\frac{\sin{\left(2 z \right)}}{z}$$$

Find $$$\int \frac{\sin{\left(2 z \right)}}{z}\, dz$$$.

Let $$$u=2 z$$$.

Then $$$du=\left(2 z\right)^{\prime }dz = 2 dz$$$ (steps can be seen »), and we have that $$$dz = \frac{du}{2}$$$.

So,

$${\color{red}{\int{\frac{\sin{\left(2 z \right)}}{z} d z}}} = {\color{red}{\int{\frac{\sin{\left(u \right)}}{u} d u}}}$$

This integral (Sine Integral) does not have a closed form:

$${\color{red}{\int{\frac{\sin{\left(u \right)}}{u} d u}}} = {\color{red}{\operatorname{Si}{\left(u \right)}}}$$

Recall that $$$u=2 z$$$:

$$\operatorname{Si}{\left({\color{red}{u}} \right)} = \operatorname{Si}{\left({\color{red}{\left(2 z\right)}} \right)}$$

Therefore,

$$\int{\frac{\sin{\left(2 z \right)}}{z} d z} = \operatorname{Si}{\left(2 z \right)}$$

Add the constant of integration:

$$\int{\frac{\sin{\left(2 z \right)}}{z} d z} = \operatorname{Si}{\left(2 z \right)}+C$$

$$$\int \frac{\sin{\left(2 z \right)}}{z}\, dz = \operatorname{Si}{\left(2 z \right)} + C$$$A