Integral of $$$\frac{\cos{\left(\frac{1}{x} \right)}}{x^{3}}$$$

Find $$$\int \frac{\cos{\left(\frac{1}{x} \right)}}{x^{3}}\, dx$$$.

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

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

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

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

For the integral $$$\int{u \cos{\left(u \right)} d u}$$$, use integration by parts $$$\int \operatorname{g} \operatorname{dv} = \operatorname{g}\operatorname{v} - \int \operatorname{v} \operatorname{dg}$$$.

Let $$$\operatorname{g}=u$$$ and $$$\operatorname{dv}=\cos{\left(u \right)} du$$$.

Then $$$\operatorname{dg}=\left(u\right)^{\prime }du=1 du$$$ (steps can be seen ») and $$$\operatorname{v}=\int{\cos{\left(u \right)} d u}=\sin{\left(u \right)}$$$ (steps can be seen »).

So,

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

The integral of the sine is $$$\int{\sin{\left(u \right)} d u} = - \cos{\left(u \right)}$$$:

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

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

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

Therefore,

$$\int{\frac{\cos{\left(\frac{1}{x} \right)}}{x^{3}} d x} = - \cos{\left(\frac{1}{x} \right)} - \frac{\sin{\left(\frac{1}{x} \right)}}{x}$$

Add the constant of integration:

$$\int{\frac{\cos{\left(\frac{1}{x} \right)}}{x^{3}} d x} = - \cos{\left(\frac{1}{x} \right)} - \frac{\sin{\left(\frac{1}{x} \right)}}{x}+C$$

$$$\int \frac{\cos{\left(\frac{1}{x} \right)}}{x^{3}}\, dx = \left(- \cos{\left(\frac{1}{x} \right)} - \frac{\sin{\left(\frac{1}{x} \right)}}{x}\right) + C$$$A