Explanation

(a)
Let $I = \int_{}^{}\frac{\cos x}{1 + \sin x}.e^{x}dx - \int_{}^{}{\frac{1}{\sin{x + 1}}.e^{x}dx}$
$= \frac{\left( \cos x \right)e^{x}}{1 + \sin x} - \int_{}^{}\left\lbrack \frac{—\left( 1 + \sin x \right)\sin x - \cos^{2}x}{\left( 1 + \sin x \right)^{2}} \right\rbrack.e^{x}dx - \int_{}^{}\frac{e^{x}}{\sin x + 1}dx$
$= \frac{e^{x}\cos x}{1 + \sin x} + \int_{}^{}\frac{1}{1 + \sin x}.e^{x}dx - \int_{}^{}\frac{e^{x}dx}{1 + \sin x} = \frac{e^{x}\cos x}{1 + \sin x} + c$