Explanation:
(D) : Given that,
\(I=\int_{0}^{\pi / 8} \cos ^{3} 4 \theta \cdot d \theta\)
\(\because \quad \cos 3 \mathrm{~A}=4 \cos ^{3} \mathrm{~A}-3 \cos \mathrm{A}\)
Then, \(\quad \cos 12 \theta=4 \cos ^{3} 4 \theta-3 \cos 4 \theta\)
\(\therefore \cos ^{3} 4 \theta=\frac{(3 \cos 4 \theta+\cos 12 \theta)}{4}\)
\(I=\int_{0}^{\pi / 8} \frac{3 \cos 4 \theta+\cos 12 \theta}{4} d \theta\)
\(I=\frac{1}{4}\left(\frac{3}{4} \sin 4 \theta+\frac{1}{12} \sin 12 \theta\right)_{0}^{\pi / 8}\)
\(=\frac{1}{4}\left(\frac{3}{4}-\frac{1}{12}\right)=\frac{1}{6}\)