Explanation

(a)
$\int_{}^{}{\frac{\cos{x - 1}}{\sin{x + 1}}e^{x}\ dx}$
$= \int_{}^{}{\frac{\cos x}{1 + \sin x}e^{x}\ dx - \int_{}^{}{\frac{1}{1 + \sin x}e^{x}\ dx}}$
$= \frac{(\cos{x)e^{x}}}{1 + \sin x} + \int_{}^{}{\frac{- (1 + \sin{x)\sin{x - \cos^{2}x}}}{(1 + {\sin{x)}}^{2}}e^{x}dx}$
$- \int_{}^{}{\frac{e^{x}}{\sin{x + 1}}dx + c}$
$= \frac{e^{x}\cos x}{1 + \sin x} + \int_{}^{}{\frac{1}{1 + \sin x}e^{x}dx - \int_{}^{}{\frac{e^{x}}{1 + \sin x}dx + c}}$
$= \frac{e^{x}\cos x}{1 + \sin x} + c$