Integral of $$$\frac{\sin{\left(x \right)} \cos{\left(x \right)}}{57}$$$

Find $$$\int \frac{\sin{\left(x \right)} \cos{\left(x \right)}}{57}\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{57}$$$ and $$$f{\left(x \right)} = \sin{\left(x \right)} \cos{\left(x \right)}$$$:

$${\color{red}{\int{\frac{\sin{\left(x \right)} \cos{\left(x \right)}}{57} d x}}} = {\color{red}{\left(\frac{\int{\sin{\left(x \right)} \cos{\left(x \right)} d x}}{57}\right)}}$$

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

The integral can be rewritten as

$$\frac{{\color{red}{\int{\sin{\left(x \right)} \cos{\left(x \right)} d x}}}}{57} = \frac{{\color{red}{\int{u d u}}}}{57}$$

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

$$\frac{{\color{red}{\int{u d u}}}}{57}=\frac{{\color{red}{\frac{u^{1 + 1}}{1 + 1}}}}{57}=\frac{{\color{red}{\left(\frac{u^{2}}{2}\right)}}}{57}$$

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

$$\frac{{\color{red}{u}}^{2}}{114} = \frac{{\color{red}{\sin{\left(x \right)}}}^{2}}{114}$$

Therefore,

$$\int{\frac{\sin{\left(x \right)} \cos{\left(x \right)}}{57} d x} = \frac{\sin^{2}{\left(x \right)}}{114}$$

Add the constant of integration:

$$\int{\frac{\sin{\left(x \right)} \cos{\left(x \right)}}{57} d x} = \frac{\sin^{2}{\left(x \right)}}{114}+C$$

$$$\int \frac{\sin{\left(x \right)} \cos{\left(x \right)}}{57}\, dx = \frac{\sin^{2}{\left(x \right)}}{114} + C$$$A