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

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

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

So,

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

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

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

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

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

Therefore,

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

Add the constant of integration:

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

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