Mind the limits !

We note that $\displaystyle \int_{0}^{\frac{\pi}{2} }\dfrac{\,dx}{1+\sin x} = 1$ . But instead I am attempting to prove that $\displaystyle \int_{0}^{\frac{\pi}{2}} \frac{\,dx}{1+ \sin x} = \infty$ . In which of the following 4 steps, I first make mistake.

Step 1: Consider the following the integral: $\displaystyle\int_{0}^{\frac{\pi}{2} }\dfrac{\,dx}{1+\sin x}$

Step 2: Rationalizing the denominator as follows: $\displaystyle\int_{0}^{\frac{\pi}{2} }\dfrac{1 -\sin x}{1-\sin^2x}\,dx$

Step 3: Now simplifying by partial decomposition: $\displaystyle\int_{0}^{\frac{\pi}{2}} \left(\sec^2 x - \tan x \sec x\right)\,dx$

Step 4: Integrating and setting the limits: $\displaystyle\int_{0}^{\frac{\pi}{2} }\dfrac{\,dx}{1+\sin x} = \infty$