Explanation

(a)
Let $I = \int_{}^{}{\frac{\sin{x + \cos x}}{3 + \sin{2x}}dx} = \int_{}^{}{\frac{\sin{x + \cos x}}{4 + \sin{2x - 1}}dx}$
$\Rightarrow \ \ \ \ \ I = \int_{}^{}\frac{\sin{x + \cos x}}{4 - \left( \sin x - \cos x \right)^{2}}dx$
Put $\sin{x - \cos x = t\ \ \ \Rightarrow \ \ \ \ \left( \cos x + \sin x \right)dx = dt}$
$\therefore\ I = \int_{}^{}\frac{dt}{4 - t^{2}} = \frac{1}{4}\log\frac{2 - t}{2 + t} + c$
$= \frac{1}{4}\log{\left( \frac{2 - \sin{x + \cos x}}{2 + \sin{x - \cos x}} \right) + c}$