Integral of $$$1 - \cos{\left(2 x \right)}$$$

Find $$$\int \left(1 - \cos{\left(2 x \right)}\right)\, dx$$$.

Integrate term by term:

$${\color{red}{\int{\left(1 - \cos{\left(2 x \right)}\right)d x}}} = {\color{red}{\left(\int{1 d x} - \int{\cos{\left(2 x \right)} d x}\right)}}$$

Apply the constant rule $$$\int c\, dx = c x$$$ with $$$c=1$$$:

$$- \int{\cos{\left(2 x \right)} d x} + {\color{red}{\int{1 d x}}} = - \int{\cos{\left(2 x \right)} d x} + {\color{red}{x}}$$

Let $$$u=2 x$$$.

Then $$$du=\left(2 x\right)^{\prime }dx = 2 dx$$$ (steps can be seen »), and we have that $$$dx = \frac{du}{2}$$$.

So,

$$x - {\color{red}{\int{\cos{\left(2 x \right)} d x}}} = x - {\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{2}$$$ and $$$f{\left(u \right)} = \cos{\left(u \right)}$$$:

$$x - {\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}} = x - {\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{2}\right)}}$$

The integral of the cosine is $$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$:

$$x - \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{2} = x - \frac{{\color{red}{\sin{\left(u \right)}}}}{2}$$

Recall that $$$u=2 x$$$:

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

Therefore,

$$\int{\left(1 - \cos{\left(2 x \right)}\right)d x} = x - \frac{\sin{\left(2 x \right)}}{2}$$

Add the constant of integration:

$$\int{\left(1 - \cos{\left(2 x \right)}\right)d x} = x - \frac{\sin{\left(2 x \right)}}{2}+C$$

$$$\int \left(1 - \cos{\left(2 x \right)}\right)\, dx = \left(x - \frac{\sin{\left(2 x \right)}}{2}\right) + C$$$A