int_{-2}^2 x^4(4-x^2)^{frac{7}{2}} dx= | vedclass

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(C) Let $x = 2 \sin 	heta$, then $dx = 2 \cos 	heta \, d	heta$. When $x = -2$, $	heta = -\frac{\pi}{2}$ and when $x = 2$, $	heta = \frac{\pi}{2}$

Vedclass · Accepted Answer

$28 \pi$

Vedclass · Answer

(C) Let $x = 2 \sin 	heta$, then $dx = 2 \cos 	heta \, d	heta$. When $x = -2$, $	heta = -\frac{\pi}{2}$ and when $x = 2$, $	heta = \frac{\pi}{2}$. Substituting these into the integral: $I = \int_{-\pi/2}^{\pi/2} (2 \sin 	heta)^4 (4 - 4 \sin^2 	heta)^{7/2} (2 \cos 	heta) \, d	heta$ $I = \int_{-\pi/2}^{\pi/2} 16 \sin^4 	heta (4 \cos^2 	heta)^{7/2} (2 \cos 	heta) \, d	heta$ $I = \int_{-\pi/2}^{\pi/2} 16 \sin^4 	heta (2^7 \cos^7 	heta) (2 \cos 	heta) \, d	heta$ $I = 16 	imes 2^8 \int_{-\pi/2}^{\pi/2} \sin^4 	heta \cos^8 	heta \, d	heta$ Since the integrand is an even function, $I = 2 	imes 2^{12} \int_{0}^{\pi/2} \sin^4 	heta \cos^8 	heta \, d	heta = 2^{13} \int_{0}^{\pi/2} \sin^4 	heta \cos^8 	heta \, d	heta$. Using Wallis' formula $\int_{0}^{\pi/2} \sin^m 	heta \cos^n 	heta \, d	heta = \frac{(m-1)!!(n-1)!!}{(m+n)!!} 	imes \frac{\pi}{2}$ (for even $m, n$): $I = 2^{13} 	imes \frac{(3 	imes 1) 	imes (7 	imes 5 	imes 3 	imes 1)}{12 	imes 10 	imes 8 	imes 6 	imes 4 	imes 2} 	imes \frac{\pi}{2}$ $I = 2^{13} 	imes \frac{3 	imes 105}{46080} 	imes \frac{\pi}{2} = 2^{13} 	imes \frac{315}{92160} 	imes \pi = 28 \pi$.

$\int_{-2}^2 x^4(4-x^2)^{\frac{7}{2}} dx=$

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