Integral of $$$\frac{\sin{\left(5 x - 3 \right)}}{t}$$$ with respect to $$$x$$$

Find $$$\int \frac{\sin{\left(5 x - 3 \right)}}{t}\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{t}$$$ and $$$f{\left(x \right)} = \sin{\left(5 x - 3 \right)}$$$:

$${\color{red}{\int{\frac{\sin{\left(5 x - 3 \right)}}{t} d x}}} = {\color{red}{\frac{\int{\sin{\left(5 x - 3 \right)} d x}}{t}}}$$

Let $$$u=5 x - 3$$$.

Then $$$du=\left(5 x - 3\right)^{\prime }dx = 5 dx$$$ (steps can be seen »), and we have that $$$dx = \frac{du}{5}$$$.

Thus,

$$\frac{{\color{red}{\int{\sin{\left(5 x - 3 \right)} d x}}}}{t} = \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{5} d u}}}}{t}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{5}$$$ and $$$f{\left(u \right)} = \sin{\left(u \right)}$$$:

$$\frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{5} d u}}}}{t} = \frac{{\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{5}\right)}}}{t}$$

The integral of the sine is $$$\int{\sin{\left(u \right)} d u} = - \cos{\left(u \right)}$$$:

$$\frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{5 t} = \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{5 t}$$

Recall that $$$u=5 x - 3$$$:

$$- \frac{\cos{\left({\color{red}{u}} \right)}}{5 t} = - \frac{\cos{\left({\color{red}{\left(5 x - 3\right)}} \right)}}{5 t}$$

Therefore,

$$\int{\frac{\sin{\left(5 x - 3 \right)}}{t} d x} = - \frac{\cos{\left(5 x - 3 \right)}}{5 t}$$

Add the constant of integration:

$$\int{\frac{\sin{\left(5 x - 3 \right)}}{t} d x} = - \frac{\cos{\left(5 x - 3 \right)}}{5 t}+C$$

$$$\int \frac{\sin{\left(5 x - 3 \right)}}{t}\, dx = - \frac{\cos{\left(5 x - 3 \right)}}{5 t} + C$$$A