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

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

Let $$$u=\frac{1}{x}$$$.

Then $$$du=\left(\frac{1}{x}\right)^{\prime }dx = - \frac{1}{x^{2}} dx$$$ (steps can be seen »), and we have that $$$\frac{dx}{x^{2}} = - du$$$.

The integral becomes

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

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

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

Rewrite the tangent as $$$\tan\left( u \right)=\frac{\sin\left( u \right)}{\cos\left( u \right)}$$$:

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

Let $$$v=\cos{\left(u \right)}$$$.

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

So,

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

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

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

The integral of $$$\frac{1}{v}$$$ is $$$\int{\frac{1}{v} d v} = \ln{\left(\left|{v}\right| \right)}$$$:

$${\color{red}{\int{\frac{1}{v} d v}}} = {\color{red}{\ln{\left(\left|{v}\right| \right)}}}$$

Recall that $$$v=\cos{\left(u \right)}$$$:

$$\ln{\left(\left|{{\color{red}{v}}}\right| \right)} = \ln{\left(\left|{{\color{red}{\cos{\left(u \right)}}}}\right| \right)}$$

Recall that $$$u=\frac{1}{x}$$$:

$$\ln{\left(\left|{\cos{\left({\color{red}{u}} \right)}}\right| \right)} = \ln{\left(\left|{\cos{\left({\color{red}{\frac{1}{x}}} \right)}}\right| \right)}$$

Therefore,

$$\int{\frac{\tan{\left(\frac{1}{x} \right)}}{x^{2}} d x} = \ln{\left(\left|{\cos{\left(\frac{1}{x} \right)}}\right| \right)}$$

Add the constant of integration:

$$\int{\frac{\tan{\left(\frac{1}{x} \right)}}{x^{2}} d x} = \ln{\left(\left|{\cos{\left(\frac{1}{x} \right)}}\right| \right)}+C$$

$$$\int \frac{\tan{\left(\frac{1}{x} \right)}}{x^{2}}\, dx = \ln\left(\left|{\cos{\left(\frac{1}{x} \right)}}\right|\right) + C$$$A