int_0^{pi/6} cos^4 3theta cdot sin^2 6theta , | vedclass

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(D) Let $I = \int_0^{\pi/6} \cos^4 3	heta \sin^2 6	heta \, d	heta$. Substitute $3	heta = t$, then $d	heta = \frac{dt}{3}$. When $	heta = 0, t =

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$\frac{5\pi}{192}$

Vedclass · Answer

(D) Let $I = \int_0^{\pi/6} \cos^4 3	heta \sin^2 6	heta \, d	heta$. Substitute $3	heta = t$, then $d	heta = \frac{dt}{3}$. When $	heta = 0, t = 0$ and when $	heta = \pi/6, t = \pi/2$. $I = \frac{1}{3} \int_0^{\pi/2} \cos^4 t \sin^2 2t \, dt$. Using $\sin 2t = 2 \sin t \cos t$, we get: $I = \frac{1}{3} \int_0^{\pi/2} \cos^4 t (2 \sin t \cos t)^2 \, dt = \frac{4}{3} \int_0^{\pi/2} \cos^6 t \sin^2 t \, dt$. Using Wallis's Formula $\int_0^{\pi/2} \sin^m x \cos^n x \, dx = \frac{(m-1)!!(n-1)!!}{(m+n)!!} \cdot \frac{\pi}{2}$ (for even $m, n$): $I = \frac{4}{3} \left[ \frac{(2-1)!!(6-1)!!}{(2+6)!!} \cdot \frac{\pi}{2} ight] = \frac{4}{3} \left[ \frac{1 \cdot 5 \cdot 3 \cdot 1}{8 \cdot 6 \cdot 4 \cdot 2} \cdot \frac{\pi}{2} ight]$. $I = \frac{4}{3} \cdot \frac{15}{384} \cdot \frac{\pi}{2} = \frac{4}{3} \cdot \frac{5}{128} \cdot \frac{\pi}{2} = \frac{5\pi}{192}$.

$\int_0^{\pi/6} \cos^4 3\theta \cdot \sin^2 6\theta \, d\theta$ is equal to

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