int_{-2 pi}^{2 pi} sin ^4 x cos ^6 x , dx = | vedclass

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Let $I = \int_{-2 \pi}^{2 \pi} \sin ^4 x \cos ^6 x \, dx$. Since $f(x) = \sin ^4 x \cos ^6 x$ is an even function,$f(-x) = \sin ^4(-x) \cos ^6(-x) = \sin ^4 x \cos ^6 x = f(x)$,we can write: $I = 2 \int_{0}^{2 \pi} \sin ^4 x \cos ^6 x \, dx$. Using the property $\int_{0}^{2a} f(x) \, dx = 4 \int_{0}^{a/2} f(x) \, dx$ if $f(2a-x) = f(x)$ and $f(a-x) = f(x)$,we note that the period of $\sin ^4 x \cos ^6 x$ is $\pi/2$. Thus,$I = 2 	imes 4 \int_{0}^{\pi/2} \sin ^4 x \cos ^6 x \, dx = 8 \int_{0}^{\pi/2} \sin ^4 x \cos ^6 x \, dx$. Using Wallis' formula $\int_{0}^{\pi/2} \sin^m x \cos^n x \, dx = \frac{(m-1)!!(n-1)!!}{(m+n)!!} 	imes \frac{\pi}{2}$ (if $m, n$ are even): $I = 8 	imes \frac{(4-1)(4-3) 	imes (6-1)(6-3)(6-5)}{(4+6)(4+6-2)(4+6-4)(4+6-6)(4+6-8)} 	imes \frac{\pi}{2}$ $I = 8 	imes \frac{3 	imes 1 	imes 5 	imes 3 	imes 1}{10 	imes 8 	imes 6 	imes 4 	imes 2} 	imes \frac{\pi}{2} = 8 	imes \frac{45}{3840} 	imes \frac{\pi}{2} = 8 	imes \frac{3}{256} 	imes \frac{\pi}{2} = \frac{24 \pi}{512} = \frac{3 \pi}{64}$.

$\int_{-2 \pi}^{2 \pi} \sin ^4 x \cos ^6 x \, dx =$

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