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

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

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,

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

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 $$$f{\left(u \right)} = \frac{1}{\cos{\left(u \right)} - 1}$$$:

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

Rewrite the cosine using the double angle formula $$$\cos\left( u \right)=1-2\sin^2\left(\frac{ u }{2}\right)$$$ and simplify:

$$- \frac{{\color{red}{\int{\frac{1}{\cos{\left(u \right)} - 1} d u}}}}{2} = - \frac{{\color{red}{\int{\left(- \frac{1}{2 \sin^{2}{\left(\frac{u}{2} \right)}}\right)d u}}}}{2}$$

Let $$$v=\frac{u}{2}$$$.

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

The integral can be rewritten as

$$- \frac{{\color{red}{\int{\left(- \frac{1}{2 \sin^{2}{\left(\frac{u}{2} \right)}}\right)d u}}}}{2} = - \frac{{\color{red}{\int{\left(- \frac{1}{\sin^{2}{\left(v \right)}}\right)d v}}}}{2}$$

Apply the constant multiple rule $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ with $$$c=-1$$$ and $$$f{\left(v \right)} = \frac{1}{\sin^{2}{\left(v \right)}}$$$:

$$- \frac{{\color{red}{\int{\left(- \frac{1}{\sin^{2}{\left(v \right)}}\right)d v}}}}{2} = - \frac{{\color{red}{\left(- \int{\frac{1}{\sin^{2}{\left(v \right)}} d v}\right)}}}{2}$$

Rewrite the integrand in terms of the cosecant:

$$\frac{{\color{red}{\int{\frac{1}{\sin^{2}{\left(v \right)}} d v}}}}{2} = \frac{{\color{red}{\int{\csc^{2}{\left(v \right)} d v}}}}{2}$$

The integral of $$$\csc^{2}{\left(v \right)}$$$ is $$$\int{\csc^{2}{\left(v \right)} d v} = - \cot{\left(v \right)}$$$:

$$\frac{{\color{red}{\int{\csc^{2}{\left(v \right)} d v}}}}{2} = \frac{{\color{red}{\left(- \cot{\left(v \right)}\right)}}}{2}$$

Recall that $$$v=\frac{u}{2}$$$:

$$- \frac{\cot{\left({\color{red}{v}} \right)}}{2} = - \frac{\cot{\left({\color{red}{\left(\frac{u}{2}\right)}} \right)}}{2}$$

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

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

Therefore,

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

Add the constant of integration:

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

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