Explanation

(a)
Let $I = \int_{\frac{- \pi}{4}}^{\frac{\pi}{4}}{\sin^{- 4}{x\ dx}} = \int_{\frac{- \pi}{4}}^{\frac{\pi}{4}}{{cosec}^{- 4}{x\ dx}}$
$= \int_{\frac{- \pi}{4}}^{\frac{\pi}{4}}{(1 + \cot^{2}x){cosec^{2}}{x\ dx}}$
Put $\cot x = t$
$\Rightarrow \ - cosec^{2}x\ dx = dt$
$\therefore\ I = - \int_{- 1}^{1}{(1 + t^{2})}dt = - 2\int_{0}^{1}{(1 + t^{2})}dt$
$= - 2\left\lbrack t + \frac{t^{3}}{3} \right\rbrack_{0}^{1} = - 2\left\lbrack 1 + \frac{1}{3} \right\rbrack = - \frac{8}{3}$