Solve: $\int \frac{\cos x}{x} dx$ Please help me solve this question. I have tried using series expansion, but I am not getting an finite answer.

There is no closed form of this integral using elementary functions,but we can show an infinite series using Taylor expansion as an answer. In case you want to know more about the integral, you can find it here
$\int \frac{\cos(x)}{x} dx=  \int( \frac{1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots}{x})dx = \int (\frac{1}{x}-\frac{x}{2!}+\frac{x^3}{4!}-\frac{x^6}{6!}+\cdots) dx= \ln(x) - \frac{x^2}{2\cdot2!}+\frac{x^4}{4\cdot4!}-\frac{x^6}{6\cdot6!}+\cdots$
So,
$\int \frac{\cos(x)}{x} dx = \ln(x) - \frac{x^2}{2\cdot2!}+\frac{x^4}{4\cdot4!}-\frac{x^6}{6\cdot6!}+\cdots$
Or more generally, $\int \frac{\cos(x)}{x} dx = \ln(x) + \Sigma_{r=1}^{\infty} (-1)^{r}\frac{x^{2r}}{2r\cdot(2r)!}$
Hope this helps !
