Integral of $$$\frac{\cos{\left(2 x \right)}}{2}$$$
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Find $$$\int \frac{\cos{\left(2 x \right)}}{2}\, dx$$$.
Solution
Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{2}$$$ and $$$f{\left(x \right)} = \cos{\left(2 x \right)}$$$:
$${\color{red}{\int{\frac{\cos{\left(2 x \right)}}{2} d x}}} = {\color{red}{\left(\frac{\int{\cos{\left(2 x \right)} d x}}{2}\right)}}$$
Let $$$u=2 x$$$.
Then $$$du=\left(2 x\right)^{\prime }dx = 2 dx$$$ (steps can be seen »), and we have that $$$dx = \frac{du}{2}$$$.
Thus,
$$\frac{{\color{red}{\int{\cos{\left(2 x \right)} d x}}}}{2} = \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}}{2}$$
Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{2}$$$ and $$$f{\left(u \right)} = \cos{\left(u \right)}$$$:
$$\frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}}{2} = \frac{{\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{2}\right)}}}{2}$$
The integral of the cosine is $$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$:
$$\frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{4} = \frac{{\color{red}{\sin{\left(u \right)}}}}{4}$$
Recall that $$$u=2 x$$$:
$$\frac{\sin{\left({\color{red}{u}} \right)}}{4} = \frac{\sin{\left({\color{red}{\left(2 x\right)}} \right)}}{4}$$
Therefore,
$$\int{\frac{\cos{\left(2 x \right)}}{2} d x} = \frac{\sin{\left(2 x \right)}}{4}$$
Add the constant of integration:
$$\int{\frac{\cos{\left(2 x \right)}}{2} d x} = \frac{\sin{\left(2 x \right)}}{4}+C$$
Answer
$$$\int \frac{\cos{\left(2 x \right)}}{2}\, dx = \frac{\sin{\left(2 x \right)}}{4} + C$$$A