calculus - Two methods of finding $int_{0}^{frac{pi}{2}} frac{dx}{1+sin x}$

Two methods of finding $$I=\int_{0}^{\pi/2} \frac{dx}{1+\sin x}$$

Method $1.$ I used substitution, $\tan \left(\frac{x}{2}\right)=t$ we get

$$I=2\int_{0}^{1}\frac{dt}{(t+1)^2}=\left. \frac{-2}{t+1}\right|_{0}^{1} =1$$

Method $2.$ we have $$I=\int_0^{\pi/2}\frac{1-\sin x}{\cos ^2 x} \,\mathrm dx = \left. \int_0^{\pi/2}\left(\sec^2 x-\sec x\tan x\right)\,\mathrm dx= \tan x-\sec x\right|_0^{\pi/2}$$

What went wrong in method $2$?