Integral of $$$- \frac{\cos{\left(93 x \right)}}{3}$$$
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Find $$$\int \left(- \frac{\cos{\left(93 x \right)}}{3}\right)\, dx$$$.
Solution
Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=- \frac{1}{3}$$$ and $$$f{\left(x \right)} = \cos{\left(93 x \right)}$$$:
$${\color{red}{\int{\left(- \frac{\cos{\left(93 x \right)}}{3}\right)d x}}} = {\color{red}{\left(- \frac{\int{\cos{\left(93 x \right)} d x}}{3}\right)}}$$
Let $$$u=93 x$$$.
Then $$$du=\left(93 x\right)^{\prime }dx = 93 dx$$$ (steps can be seen »), and we have that $$$dx = \frac{du}{93}$$$.
Thus,
$$- \frac{{\color{red}{\int{\cos{\left(93 x \right)} d x}}}}{3} = - \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{93} d u}}}}{3}$$
Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{93}$$$ and $$$f{\left(u \right)} = \cos{\left(u \right)}$$$:
$$- \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{93} d u}}}}{3} = - \frac{{\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{93}\right)}}}{3}$$
The integral of the cosine is $$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$:
$$- \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{279} = - \frac{{\color{red}{\sin{\left(u \right)}}}}{279}$$
Recall that $$$u=93 x$$$:
$$- \frac{\sin{\left({\color{red}{u}} \right)}}{279} = - \frac{\sin{\left({\color{red}{\left(93 x\right)}} \right)}}{279}$$
Therefore,
$$\int{\left(- \frac{\cos{\left(93 x \right)}}{3}\right)d x} = - \frac{\sin{\left(93 x \right)}}{279}$$
Add the constant of integration:
$$\int{\left(- \frac{\cos{\left(93 x \right)}}{3}\right)d x} = - \frac{\sin{\left(93 x \right)}}{279}+C$$
Answer
$$$\int \left(- \frac{\cos{\left(93 x \right)}}{3}\right)\, dx = - \frac{\sin{\left(93 x \right)}}{279} + C$$$A