Explanation

(a)
Let $I = \int_{0}^{\pi/2}{\frac{\cos x}{(1 + \sin{x)(2 + \sin{x)}}}dx}$
Put $\sin{x = t\ \Rightarrow \ \cos{x\ dx = dt}}$
$\therefore\ I = \int_{0}^{1}\frac{dt}{(1 + t)(2 + t)}$
$= \int_{0}^{1}{\frac{1}{(1 + t)}dt - \int_{0}^{1}{\frac{1}{(2 + 1)}dt}}$
$= \lbrack\log(1 + t) - \log{(2 + t)\rbrack}_{0}^{1}$
$= (\log{2 - \log{3) - (0 - \log{2)}}}$
$= \log\frac{4}{3}$