Integral of $$$\cot^{2}{\left(\frac{x}{2} \right)} \csc^{2}{\left(\frac{x}{2} \right)}$$$

Find $$$\int \cot^{2}{\left(\frac{x}{2} \right)} \csc^{2}{\left(\frac{x}{2} \right)}\, dx$$$.

Let $$$u=\cot{\left(\frac{x}{2} \right)}$$$.

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

The integral can be rewritten as

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

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

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

Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=2$$$:

$$- 2 {\color{red}{\int{u^{2} d u}}}=- 2 {\color{red}{\frac{u^{1 + 2}}{1 + 2}}}=- 2 {\color{red}{\left(\frac{u^{3}}{3}\right)}}$$

Recall that $$$u=\cot{\left(\frac{x}{2} \right)}$$$:

$$- \frac{2 {\color{red}{u}}^{3}}{3} = - \frac{2 {\color{red}{\cot{\left(\frac{x}{2} \right)}}}^{3}}{3}$$

Therefore,

$$\int{\cot^{2}{\left(\frac{x}{2} \right)} \csc^{2}{\left(\frac{x}{2} \right)} d x} = - \frac{2 \cot^{3}{\left(\frac{x}{2} \right)}}{3}$$

Add the constant of integration:

$$\int{\cot^{2}{\left(\frac{x}{2} \right)} \csc^{2}{\left(\frac{x}{2} \right)} d x} = - \frac{2 \cot^{3}{\left(\frac{x}{2} \right)}}{3}+C$$

$$$\int \cot^{2}{\left(\frac{x}{2} \right)} \csc^{2}{\left(\frac{x}{2} \right)}\, dx = - \frac{2 \cot^{3}{\left(\frac{x}{2} \right)}}{3} + C$$$A