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

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(A) माना $I = \int_{-2 \pi}^{2 \pi} \sin ^4(2 x) \cos ^6(2 x) d x$. चूंकि $f(x) = \sin ^4(2 x) \cos ^6(2 x)$ एक सम फलन है,इसलिए $I = 2 \int_{0}^{2 \pi

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) माना $I = \int_{-2 \pi}^{2 \pi} \sin ^4(2 x) \cos ^6(2 x) d x$. चूंकि $f(x) = \sin ^4(2 x) \cos ^6(2 x)$ एक सम फलन है,इसलिए $I = 2 \int_{0}^{2 \pi} \sin ^4(2 x) \cos ^6(2 x) d x$. माना $2x = t$,तब $2 dx = dt$,अर्थात $dx = \frac{1}{2} dt$. जब $x = 0, t = 0$ और जब $x = 2 \pi, t = 4 \pi$. $I = 2 \int_{0}^{4 \pi} \sin ^4(t) \cos ^6(t) \frac{1}{2} dt = \int_{0}^{4 \pi} \sin ^4(t) \cos ^6(t) dt$. गुणधर्म $\int_{0}^{n T} f(t) dt = n \int_{0}^{T} f(t) dt$ का उपयोग करते हुए,जहाँ $T = \pi$: $I = 4 \int_{0}^{\pi} \sin ^4(t) \cos ^6(t) dt = 4 	imes 2 \int_{0}^{\pi/2} \sin ^4(t) \cos ^6(t) dt = 8 \int_{0}^{\pi/2} \sin ^4(t) \cos ^6(t) dt$. वालिस के सूत्र $\int_{0}^{\pi/2} \sin^m(x) \cos^n(x) dx = \frac{(m-1)!!(n-1)!!}{(m+n)!!} 	imes \frac{\pi}{2}$ का उपयोग करते हुए: $I = 8 	imes \frac{(3 	imes 1) 	imes (5 	imes 3 	imes 1)}{(10 	imes 8 	imes 6 	imes 4 	imes 2)} 	imes \frac{\pi}{2} = 8 	imes \frac{45 \pi}{7680} = \frac{3 \pi}{64}$.

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

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