Hello, fardeen_gen!
Derive formula of lateral (curved) surface area of cone using integral calculus. I read that proof, and don't care for it.
I think those "triangles" are very clumsy to handle.
I would approach it like this . . . Code:

r + * (h,r)
 * :
 * :
 * :
  *        + 
 h
We have a right triangle formed by the xaxis, $\displaystyle y \:=\:\frac{r}{h}x$, and $\displaystyle x = h$
Rotate the triangle about the xaxis to form a cone with radius $\displaystyle r$ and height $\displaystyle h.$
The surface area is given by: .$\displaystyle S \;=\;2\pi\int^a_b y\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx $
So we have: .$\displaystyle S \;=\;2\pi\int^h_0\left(\frac{r}{h}x\right)\sqrt{1 + \left(\frac{r}{h}\right)^2}\,dx \;=\;\frac{2\pi r\sqrt{r^2+h^2}}{h^2}\int^h_0 x\,dx $
. . . . . . . $\displaystyle = \;\frac{2\pi r\sqrt{r^2+h^2}}{h^2}\cdot\frac{1}{2}x^2\,\bigg]^h_0 \;=\;\frac{2\pi r\sqrt{r^2+h^2}}{h^2}\cdot \frac{1}{2}h^2 $
Therefore: .$\displaystyle S \;=\;\pi r\sqrt{r^2+h^2}$