Explanation

(b)
Let $I = \int_{}^{}{\sin{x\ d(\cos{x)}}}$
Put $\cos{x = t\ \Rightarrow d(\cos{x) = dt}}$
$\therefore I = \int_{}^{}{\sqrt{1 - t^{2}}dt}$
$= \frac{t}{2}\sqrt{1 - t^{2}} + \frac{1}{2}\sin^{- 1}{(t) + c}$
$= \frac{\cos x}{2} \bullet \sin{x + \frac{1}{2}\sin^{- 1}{(\cos{x) + c}}}$
$= \frac{1}{4}\sin{2x + \frac{1}{2}\sin^{- 1}{\left\lbrack \sin\left( \frac{\pi}{2} - x\ \right) \right\rbrack + c}}$
$= \frac{1}{4}\sin{2x - \frac{x}{2} + c}$