Explanation

(d)
Let $I = \int_{0}^{\pi/8}{\cos^{3}{4\theta\ d\theta}}$
$= \int_{0}^{\pi/8}{\cos^{2}{4\theta \bullet \cos{4\theta}\ d\theta}}$
$= \int_{0}^{\pi/8}{\left( \frac{1 + \cos{8\theta}}{2} \right)\cos{4\theta\ d\theta}}$
$= \frac{1}{2}\int_{0}^{\pi/8}{\cos{4\theta\ d\theta}} + \frac{1}{2}\int_{0}^{\pi/8}{{\cos\ 8\theta}{\cos{4\theta\ d\theta}}}$
$= \frac{1}{2}\left\lbrack \frac{\sin{4\theta}}{4} \right\rbrack_{0}^{\pi/8} + I_{1}$
$I = \frac{1}{8} + I_{1}$
Where, $I_{1} = \frac{1}{2}\int_{0}^{\pi/8}{{\cos\ 8\theta}{\cos{4\theta\ d\theta}}}$
$= \frac{1}{4}\int_{0}^{\pi/8}{{2\ cos\ 8\theta}{\cos{4\theta\ d\theta}}}$
$= \frac{1}{4}\left\lbrack \int_{0}^{\pi/8}{\cos{12\theta\ d\theta} + \int_{0}^{\pi/8}{\cos{4\theta\ d\theta}}} \right\rbrack$
$= \frac{1}{4}\left\lbrack \frac{\sin{12\theta}}{12} + \frac{\sin{4\theta}}{4} \right\rbrack_{0}^{\pi/8}$
$= \frac{1}{4}\left\lbrack - \frac{1}{21} + \frac{1}{4} \right\rbrack = \frac{1}{4}\left\lbrack \frac{2}{12} \right\rbrack 0$
$I_{1} = \frac{1}{24}$ …(iii)
From Eqs.(i) and (ii), we get
$I = \frac{1}{8} + \frac{1}{24} - \frac{3 + 1}{24} = \frac{1}{6}$