Explanation

(a)
Let $I = \int_{}^{}\frac{dx}{(\sin{x - 2\cos{x)(2\sin{x + \cos{x)}}}}}$
$= \int_{}^{}\frac{\sec^{2}{x\ dx}}{(\tan{x - 2)(2\tan{x + 1)}}}$
Put $\tan{x = t \Rightarrow \sec^{2}{x\ dx = dt}}$
$\therefore I = \int_{}^{}\frac{dt}{(t - 2)(2t + 1)}$
$= \frac{1}{5}\int_{}^{}{\frac{t}{t - 2}dt - \frac{2}{5}\int_{}^{}{\frac{1}{2t + 1}dt}}$
$= \frac{1}{5}\log{(t - 2) - \frac{1}{5}\log{(2t + 1) + c}}$
$= \log{\sqrt[5]{\frac{\tan{x - 2}}{2\tan{x + 1}}} + c}$