Integral of $$$\sin{\left(\frac{\pi x}{3} \right)} \cos{\left(\frac{\pi x}{3} \right)}$$$

Find $$$\int \sin{\left(\frac{\pi x}{3} \right)} \cos{\left(\frac{\pi x}{3} \right)}\, dx$$$.

Let $$$u=\sin{\left(\frac{\pi x}{3} \right)}$$$.

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

Therefore,

$${\color{red}{\int{\sin{\left(\frac{\pi x}{3} \right)} \cos{\left(\frac{\pi x}{3} \right)} d x}}} = {\color{red}{\int{\frac{3 u}{\pi} d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{3}{\pi}$$$ and $$$f{\left(u \right)} = u$$$:

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

Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=1$$$:

$$\frac{3 {\color{red}{\int{u d u}}}}{\pi}=\frac{3 {\color{red}{\frac{u^{1 + 1}}{1 + 1}}}}{\pi}=\frac{3 {\color{red}{\left(\frac{u^{2}}{2}\right)}}}{\pi}$$

Recall that $$$u=\sin{\left(\frac{\pi x}{3} \right)}$$$:

$$\frac{3 {\color{red}{u}}^{2}}{2 \pi} = \frac{3 {\color{red}{\sin{\left(\frac{\pi x}{3} \right)}}}^{2}}{2 \pi}$$

Therefore,

$$\int{\sin{\left(\frac{\pi x}{3} \right)} \cos{\left(\frac{\pi x}{3} \right)} d x} = \frac{3 \sin^{2}{\left(\frac{\pi x}{3} \right)}}{2 \pi}$$

Add the constant of integration:

$$\int{\sin{\left(\frac{\pi x}{3} \right)} \cos{\left(\frac{\pi x}{3} \right)} d x} = \frac{3 \sin^{2}{\left(\frac{\pi x}{3} \right)}}{2 \pi}+C$$

$$$\int \sin{\left(\frac{\pi x}{3} \right)} \cos{\left(\frac{\pi x}{3} \right)}\, dx = \frac{3 \sin^{2}{\left(\frac{\pi x}{3} \right)}}{2 \pi} + C$$$A