int_{-2 pi}^{2 pi} sin ^4(2 x) cos ^6(2 x) d | vedclass

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Question

(A) Let $I = \int_{-2 \pi}^{2 \pi} \sin ^4(2 x) \cos ^6(2 x) d x$. Since $f(x) = \sin ^4(2 x) \cos ^6(2 x)$ is an even function,$I = 2 \int_{0}^{2 \pi

Vedclass · Accepted Answer

$\frac{3 \pi}{64}$

Vedclass · Answer

(A) Let $I = \int_{-2 \pi}^{2 \pi} \sin ^4(2 x) \cos ^6(2 x) d x$. Since $f(x) = \sin ^4(2 x) \cos ^6(2 x)$ is an even function,$I = 2 \int_{0}^{2 \pi} \sin ^4(2 x) \cos ^6(2 x) d x$. Let $2x = t$,then $2 dx = dt$,so $dx = \frac{1}{2} dt$. When $x = 0, t = 0$ and when $x = 2 \pi, t = 4 \pi$. $I = 2 \int_{0}^{4 \pi} \sin ^4(t) \cos ^6(t) \frac{1}{2} dt = \int_{0}^{4 \pi} \sin ^4(t) \cos ^6(t) dt$. Using the property $\int_{0}^{n T} f(t) dt = n \int_{0}^{T} f(t) dt$,where $T = \pi$: $I = 4 \int_{0}^{\pi} \sin ^4(t) \cos ^6(t) dt = 4 	imes 2 \int_{0}^{\pi/2} \sin ^4(t) \cos ^6(t) dt = 8 \int_{0}^{\pi/2} \sin ^4(t) \cos ^6(t) dt$. Using Wallis' formula $\int_{0}^{\pi/2} \sin^m(x) \cos^n(x) dx = \frac{(m-1)!!(n-1)!!}{(m+n)!!}$: $I = 8 	imes \frac{(3 	imes 1) 	imes (5 	imes 3 	imes 1)}{(10 	imes 8 	imes 6 	imes 4 	imes 2)} = 8 	imes \frac{3 	imes 15}{3840} = 8 	imes \frac{45}{3840} = \frac{45}{480} = \frac{3}{32}$. Wait,re-evaluating the integral: $\int_{0}^{\pi/2} \sin^4(t) \cos^6(t) dt = \frac{3 \cdot 1 \cdot 5 \cdot 3 \cdot 1}{10 \cdot 8 \cdot 6 \cdot 4 \cdot 2} \cdot \frac{\pi}{2} = \frac{45 \pi}{7680} = \frac{3 \pi}{512}$. $I = 8 	imes \frac{3 \pi}{512} = \frac{3 \pi}{64}$.

$\int_{-2 \pi}^{2 \pi} \sin ^4(2 x) \cos ^6(2 x) d x=$

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