$\int \mathrm{tg}xdx=$

Step 1 ：$\int \mathrm{tg}xdx=\int \frac{\sin x}{\cos x}dx$

Step 2 ：Let $u=\sin x$ and $dv=\frac{1}{\cos x}dx$

Step 3 ：Then, $du=\cos xdx$ and $v=\int \frac{1}{\cos x}dx$

Step 4 ：Let $y=\tan (\frac{x}{2})$, then $\cos x=\frac{1-y^2}{1+y^2}$ and $dx=\frac{2}{1+y^2}dy$

Step 5 ：$\int \frac{1}{\cos x}dx=\int \frac{1+y^2}{1-y^2}\cdot \frac{2}{1+y^2}dy=\int \frac{2}{1-y^2}dy$

Step 6 ：$\int \frac{2}{1-y^2}dy=-\ln (y-1)+\ln (y+1)$

Step 7 ：Substitute back for $x$: $v=-\ln (\tan (\frac{x}{2})-1)+\ln (\tan (\frac{x}{2})+1)$

Step 8 ：Using integration by parts: $\int \mathrm{tg}xdx=uv-\int vdu$

Step 9 ：$\int \mathrm{tg}xdx=\sin x(-\ln (\tan (\frac{x}{2})-1)+\ln (\tan (\frac{x}{2})+1))-\int (-\ln (\tan (\frac{x}{2})-1)+\ln (\tan (\frac{x}{2})+1))\cos xdx$

Step 10 ：$\int \mathrm{tg}xdx=\ln (\cos x)+C$

Step 11 ：$\boxed{{\displaystyle \int \mathrm{tg}xdx=\ln (\cos x)+C}}$