Evaluate the following integrals:

Assume $\frac{1}{x}=t$

$\Rightarrow \frac{1}{\mathrm{x}^{2}} \mathrm{dx}=\mathrm{dt}$

Substituting $t$ and dt we get

$\Rightarrow \int \cos ^{2} t d t$

$\therefore$ The given equation becomes,

$\Rightarrow \int \frac{1-\cos 2 t}{2} \mathrm{dx}$

We know $\int \cos a x d x=\frac{1}{a} \sin a x+c$

$\Rightarrow \frac{1}{2} \int \mathrm{d} x t-\frac{1}{2} \int \cos (2 t) d t$

$\Rightarrow \frac{t}{2}-\frac{1}{4} \sin (t)+c$

But $\frac{1}{x}=t$

$\Rightarrow \frac{1}{2 x}-\frac{1}{4} \sin \left(\frac{1}{x}\right)+c$