# Need help understanding $\frac {dx}{\cos^2(\frac{x}{2})} = 2d(\operatorname{tg}(\frac{x}{2}))$

I have found this statement somewhere, however, I dont really understand it.

Could someone explain me where does $2$ before $\operatorname{tg}(x/2)$ come from?

$$\frac {dx}{\cos^2(\frac{x}{2})} = 2d(\operatorname{tg}(\frac{x}{2}))$$

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$\displaystyle{sec^2 x = \frac{1}{cos^2 x}}$ –  Kirthi Raman Mar 15 '12 at 2:49

• $d(\tan x)=\sec^2 x \rm dx$
• $\operatorname{tg}(x)$ is an archaic name for $\tan x$
$d(\tan \dfrac x 2)=\dfrac 1 2\sec^2\dfrac x 2 \mathrm{d}x \implies \dfrac{\mathrm{d}x}{\cos^2 \dfrac{x}{2}}=2\mathrm{d}(\tan \dfrac{x}{2})$
$\mathrm{tg}$ is also the notation used in Russia and the Balkan countries, if memory serves. –  Ｊ. Ｍ. Aug 1 '12 at 8:16