Integral of $$$\cos{\left(3 x \right)}$$$
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Solution
Let $$$u=3 x$$$.
Then $$$du=\left(3 x\right)^{\prime }dx = 3 dx$$$ (steps can be seen »), and we have that $$$dx = \frac{du}{3}$$$.
Therefore,
$${\color{red}{\int{\cos{\left(3 x \right)} d x}}} = {\color{red}{\int{\frac{\cos{\left(u \right)}}{3} d u}}}$$
Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{3}$$$ and $$$f{\left(u \right)} = \cos{\left(u \right)}$$$:
$${\color{red}{\int{\frac{\cos{\left(u \right)}}{3} d u}}} = {\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{3}\right)}}$$
The integral of the cosine is $$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$:
$$\frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{3} = \frac{{\color{red}{\sin{\left(u \right)}}}}{3}$$
Recall that $$$u=3 x$$$:
$$\frac{\sin{\left({\color{red}{u}} \right)}}{3} = \frac{\sin{\left({\color{red}{\left(3 x\right)}} \right)}}{3}$$
Therefore,
$$\int{\cos{\left(3 x \right)} d x} = \frac{\sin{\left(3 x \right)}}{3}$$
Add the constant of integration:
$$\int{\cos{\left(3 x \right)} d x} = \frac{\sin{\left(3 x \right)}}{3}+C$$
Answer
$$$\int \cos{\left(3 x \right)}\, dx = \frac{\sin{\left(3 x \right)}}{3} + C$$$A