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**Prove It** $\displaystyle \displaystyle \begin{align*} \int{\frac{1 + \sin{x}}{\cos{x}}\,dx} &= \int{\frac{(1 + \sin{x})(1 - \sin{x})}{\cos{x}(1 - \sin{x})}\,dx} \\ &= \int{\frac{1 - \sin^2{x}}{\cos{x}(1 - \sin{x})}\,dx} \\ &= \int{\frac{\cos^2{x}}{\cos{x}(1 - \sin{x})}\,dx} \\ &= \int{\frac{\cos{x}}{1 - \sin{x}}\,dx} \\ &= -\int{\frac{-\cos{x}}{1 - \sin{x}}\,dx} \\ &= -\int{\frac{1}{u}\,du}\textrm{ after making the substitution }u = 1 - \sin{x} \implies du = -\cos{x}\,dx\end{align*}$

I'm sure you can go from here...