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

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

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

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

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 $$$\cos{\left(x \right)} dx = du$$$.

Thus,

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

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

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

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

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

Therefore,

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

Add the constant of integration:

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

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