Evaluate the following integrals:

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

$\Rightarrow \frac{1}{x^{2}} d x=d t$

Substituting $\mathrm{t}$ and $\mathrm{dt}$ we get

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

$\Rightarrow \cos ^{2} x=\frac{1+\cos 2 x}{2}$

$\therefore$ The given equation becomes,

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

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

$\Rightarrow \frac{1}{2} \int \mathrm{dxt}-\frac{1}{2} \int \cos (2 \mathrm{t}) \mathrm{dt}$

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

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

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