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

Find $$$\int \frac{\sin{\left(\pi x \right)}}{2}\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{2}$$$ and $$$f{\left(x \right)} = \sin{\left(\pi x \right)}$$$:

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

Let $$$u=\pi x$$$.

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

So,

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

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

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

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

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

Recall that $$$u=\pi x$$$:

$$- \frac{\cos{\left({\color{red}{u}} \right)}}{2 \pi} = - \frac{\cos{\left({\color{red}{\pi x}} \right)}}{2 \pi}$$

Therefore,

$$\int{\frac{\sin{\left(\pi x \right)}}{2} d x} = - \frac{\cos{\left(\pi x \right)}}{2 \pi}$$

Add the constant of integration:

$$\int{\frac{\sin{\left(\pi x \right)}}{2} d x} = - \frac{\cos{\left(\pi x \right)}}{2 \pi}+C$$

$$$\int \frac{\sin{\left(\pi x \right)}}{2}\, dx = - \frac{\cos{\left(\pi x \right)}}{2 \pi} + C$$$A