I know of the conventional ways of finding secant integral through logarithmic derivative, however I was trying to solve it through other method and don't understand why it doesn't work.
So
$$\int \sec(x)dx = \int\frac{1}{\cos(x)}dx = \int\frac{\cos^2(x)+\sin^2(x)}{\cos(x)}dx $$$$= \int\cos(x)dx+\int\frac{\sin^2(x)}{\cos(x)}dx = \int\cos(x)dx + \int \tan(x)\sec(x)dx$$
$\cos(x)$ is easy to integrate and since $$\frac{d\sec(x)}{dx} = \tan(x)\sec(x)$$ using FTC 2, we get
$$\sin(x)+\sec(x) + C, \, C\in \Bbb R$$
Can somebody explain where does my calculation start going wrong?
REMARK
Here it is your mistake:
\begin{align*} \int\frac{\sin^{2}(x)}{\cos(x)}\mathrm{d}x = \int\tan(x)\sec(x)\mathrm{d}x \end{align*}
This is another approach I find interesting. Substitute $u = \sin(x)$ in order to obtain \begin{align*} \int\sec(x)\mathrm{d}x & = \int\frac{\mathrm{d}x}{\cos(x)} = \int\frac{\cos(x)}{\cos^{2}(x)}\mathrm{d}x = \int\frac{\cos(x)}{1-\sin^{2}(x)}\mathrm{d}x\\\\ & = \int\frac{\mathrm{d}u}{1-u^{2}} = \frac{1}{2}\left[\int\frac{\mathrm{d}u}{1+u} + \int\frac{\mathrm{d}u}{1-u}\right]\\\\ & = \frac{1}{2}\left[\ln\left|1+u\right| - \ln|1-u|\right] = \frac{1}{2}\ln\left|\frac{1+u}{1-u}\right| = \frac{1}{2}\ln\left|\frac{1+\sin(x)}{1-\sin(x)}\right|\\\\ & = \frac{1}{2}\ln\left|\frac{\sec(x) + \tan(x)}{\sec(x) - \tan(x)}\right| = \frac{1}{2}\ln\left|\frac{(\sec(x) + \tan(x))^{2}}{\sec^{2}(x) - \tan^{2}(x)}\right| = \ln\left|\sec(x)+\tan(x)\right| \end{align*}