Integral of $$$\sin{\left(\frac{3 u}{5} \right)}$$$

Find $$$\int \sin{\left(\frac{3 u}{5} \right)}\, du$$$.

Let $$$v=\frac{3 u}{5}$$$.

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

The integral can be rewritten as

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

Apply the constant multiple rule $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ with $$$c=\frac{5}{3}$$$ and $$$f{\left(v \right)} = \sin{\left(v \right)}$$$:

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

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

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

Recall that $$$v=\frac{3 u}{5}$$$:

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

Therefore,

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

Add the constant of integration:

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

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