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

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

Expand the expression:

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

Integrate term by term:

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

Rewrite the integrand in terms of the cosecant:

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

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

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

Let $$$u=\sin{\left(x \right)}$$$.

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

The integral becomes

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

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

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

Recall that $$$u=\sin{\left(x \right)}$$$:

$$- \cot{\left(x \right)} + {\color{red}{u}}^{-1} = - \cot{\left(x \right)} + {\color{red}{\sin{\left(x \right)}}}^{-1}$$

Therefore,

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

Add the constant of integration:

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

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