int_0^{pi /8} cos^3(4theta) , dtheta = | vedclass

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(D) Let $I = \int_0^{\pi /8} \cos^3(4	heta) \, d	heta = \int_0^{\pi /8} \cos^2(4	heta) \cos(4	heta) \, d	heta$. Using the identity $\cos^2(4	het

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$\frac{1}{6}$

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(D) Let $I = \int_0^{\pi /8} \cos^3(4	heta) \, d	heta = \int_0^{\pi /8} \cos^2(4	heta) \cos(4	heta) \, d	heta$. Using the identity $\cos^2(4	heta) = 1 - \sin^2(4	heta)$,we get: $I = \int_0^{\pi /8} (1 - \sin^2(4	heta)) \cos(4	heta) \, d	heta$. Let $t = \sin(4	heta)$. Then $dt = 4 \cos(4	heta) \, d	heta$,which implies $\cos(4	heta) \, d	heta = \frac{dt}{4}$. When $	heta = 0$,$t = \sin(0) = 0$. When $	heta = \frac{\pi}{8}$,$t = \sin(4 	imes \frac{\pi}{8}) = \sin(\frac{\pi}{2}) = 1$. Substituting these into the integral: $I = \frac{1}{4} \int_0^1 (1 - t^2) \, dt = \frac{1}{4} [t - \frac{t^3}{3}]_0^1$. $I = \frac{1}{4} (1 - \frac{1}{3}) = \frac{1}{4} (\frac{2}{3}) = \frac{1}{6}$.

$\int_0^{\pi /8} \cos^3(4\theta) \, d\theta = $

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