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

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

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

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

Rewrite in terms of the cotangent:

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

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

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

Therefore,

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

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

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

Rewrite and split the fraction:

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

Integrate term by term:

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

Apply the constant rule $$$\int c\, du = c u$$$ with $$$c=1$$$:

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

The integral of $$$\frac{1}{u^{2} + 1}$$$ is $$$\int{\frac{1}{u^{2} + 1} d u} = \operatorname{atan}{\left(u \right)}$$$:

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

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

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

Therefore,

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

Simplify:

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

Add the constant of integration:

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

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